主席树区间修改
#define IO std::ios::sync_with_stdio(0),cin.tie(0),cout.tie(0)
#define bug(x) cout<<#x<<" is "<<x<<endl
#include <bits/stdc++.h>
#define iter ::iterator
using namespace std;
typedef long long ll;
typedef pair<int,int>P;
#define pb push_back
#define mk make_pair
#define se second
#define fi first
const ll mod=1e9+7;
const int N=1e5+5;
int n,q;
int a[N],rt[N*32],ls[N*32],rs[N*32];
ll sum[N*32],add[N*32];
int cnt;
void build(int &o,int l,int r){
o=++cnt;
if(l==r){
sum[o]=a[l];
return;
}
int m=(l+r)/2;
build(ls[o],l,m);
build(rs[o],m+1,r);
sum[o]=sum[ls[o]]+sum[rs[o]];
}
void up(int &o,int pre,int l,int r,int ql,int qr,int val){
o=++cnt;
ls[o]=ls[pre];
rs[o]=rs[pre];
add[o]=add[pre];
sum[o]=sum[pre]+1ll*val*(min(r,qr)-max(l,ql)+1);
if(l>=ql&&r<=qr){
add[o]+=val;
return;
}
int m=(l+r)/2;
if(ql<=m)up(ls[o],ls[pre],l,m,ql,qr,val);
if(qr>m)up(rs[o],rs[pre],m+1,r,ql,qr,val);
}
ll qu(int o,int l,int r,int ql,int qr){
if(l>=ql&&r<=qr)return sum[o];
ll res=add[o]*(min(r,qr)-max(l,ql)+1);
int m=(l+r)/2;
if(ql<=m)res+=qu(ls[o],l,m,ql,qr);
if(qr>m)res+=qu(rs[o],m+1,r,ql,qr);
return res;
}
int main(){
while(~scanf("%d%d",&n,&q)){
for(int i=1;i<=n;i++)scanf("%d",&a[i]);
cnt=0;
build(rt[0],1,n);
int time=0;
while(q--){
char op;
int L,R,val;
scanf(" %c",&op);
if(op=='Q'){
scanf("%d%d",&L,&R);
ll ans=qu(rt[time],1,n,L,R);
printf("%lld\n",ans);
}
else if(op=='C'){
scanf("%d%d%d",&L,&R,&val);
++time;
up(rt[time],rt[time-1],1,n,L,R,val);
}
else if(op=='H'){
scanf("%d%d%d",&L,&R,&val);
ll ans=qu(rt[val],1,n,L,R);
printf("%lld\n",ans);
}
else{
scanf("%d",&val);
time=val;
}
}
printf("\n");
}
}
带权并查集
#include <bits/stdc++.h>
using namespace std;
const int N = 200010;
int parent[N];
int value[N]; ///记录当前结点到根结点的距离
int FindSet(int x) {
if (x == parent[x])
return x;
else {
int t = parent[x]; //记录原父节点编号
parent[x] = FindSet(parent[x]); //父节点变为根节点，此时value[x]=父节点到根节点的权值
value[x] += value[t]; //当前节点的权值加上原本父节点的权值
return parent[x];
}
}
int main(){
int n, m;
while (scanf("%d%d", &n, &m) != EOF) {
for (int i = 0; i <= n; i++) {
parent[i] = i;
value[i] = 0;
}
int ans = 0;
while (m--) {
int a, b, v;
scanf("%d%d%d", &a, &b, &v);
a--;
int roota = FindSet(a);
int rootb = FindSet(b);
if (roota == rootb) {
if (value[a] - value[b] != v) ans++; ///精华部分1
}
else {
parent[roota] = rootb;
value[roota] = -value[a] + value[b] + v; ///精华部分2
}
}
printf("%d\n", ans);
}
}
树剖
#define IO std::ios::sync_with_stdio(0);
#include <bits/stdc++.h>
#define iter ::iterator
using namespace std;
typedef long long ll;
typedef pair<ll,ll>P;
#define pb push_back
#define se second
#define fi first
#define rs o<<1|1
#define ls o<<1
#define inf 0x3f3f3f3f
const int N=1e5+5;
int n,q,rt,mod;
int a[N],b[N];
vector<int>g[N];
int deep[N],fa[N],id[N],son[N],siz[N],top[N];
int sumv[N*4],add[N*4];
int L[N*4],R[N*4];
void dfs1(int u,int f){
deep[u]=deep[f]+1;
fa[u]=f;
siz[u]=1;
int mx=-1;
for(int i=0;i<g[u].size();i++){
int v=g[u][i];
if(v==f)continue;
dfs1(v,u);
siz[u]+=siz[v];
if(siz[v]>mx){
mx=siz[v];
son[u]=v;
}
}
}
int tim;
void dfs2(int u,int topf){
id[u]=++tim;
a[tim]=b[u];
top[u]=topf;
if(!son[u])return;
dfs2(son[u],topf);
for(int i=0;i<g[u].size();i++){
int v=g[u][i];
if(v==fa[u]||v==son[u])continue;
dfs2(v,v);
}
}
void push(int o){
sumv[o]=(sumv[ls]+sumv[rs])%mod;
}
void down(int o){
if(add[o]){
add[o]%=mod;
sumv[ls]+=add[o]*(R[ls]-L[ls]+1);
sumv[ls]%=mod;
sumv[rs]+=add[o]*(R[rs]-L[rs]+1);
sumv[rs]%=mod;
add[ls]+=add[o];
add[ls]%=mod;
add[rs]+=add[o];
add[rs]%=mod;
add[o]=0;
}
}
void build(int o,int l,int r){
if(l==r){
sumv[o]=a[l]%mod;
L[o]=R[o]=l;
return;
}
int m=(l+r)/2;
build(ls,l,m);
build(rs,m+1,r);
L[o]=L[ls];
R[o]=R[rs];
push(o);
}
int qu(int o,int l,int r,int ql,int qr){
if(l>=ql&&r<=qr){
return sumv[o]%mod;
}
down(o);
int m=(l+r)/2;
int res=0;
if(ql<=m)res+=qu(ls,l,m,ql,qr)%mod;
res%=mod;
if(qr>m)res+=qu(rs,m+1,r,ql,qr)%mod;
res%=mod;
return res;
}
void up(int o,int l,int r,int ql,int qr,int v){
if(l>=ql&&r<=qr){
v%=mod;
add[o]+=v;
add[o]%=mod;
sumv[o]+=v*(r-l+1);
sumv[o]%=mod;
return;
}
down(o);
int m=(l+r)/2;
if(ql<=m)up(ls,l,m,ql,qr,v);
if(qr>m)up(rs,m+1,r,ql,qr,v);
push(o);
}
void qu1(int x,int y){//x节点到y节点的路径和
int res=0;
while(top[x]!=top[y]){
if(deep[top[x]]<deep[top[y]])swap(x,y);
res+=qu(1,1,n,id[top[x]],id[x]);
res%=mod;
x=fa[top[x]];
}
if(deep[x]>deep[y])swap(x,y);
res+=qu(1,1,n,id[x],id[y])%mod;
res%=mod;
printf("%d\n",res);
}
void up1(int x,int y,int z){//把x节点到y节点的路径所有点加z
while(top[x]!=top[y]){
if(deep[top[x]]<deep[top[y]])swap(x,y);
up(1,1,n,id[top[x]],id[x],z);
x=fa[top[x]];
}
if(deep[x]>deep[y])swap(x,y);
up(1,1,n,id[x],id[y],z);
}
int main(){
scanf("%d%d%d%d",&n,&q,&rt,&mod);
for(int i=1;i<=n;i++){
scanf("%d",&b[i]);
b[i]%=mod;
}
for(int i=1;i<n;i++){
int x,y;
scanf("%d%d",&x,&y);
g[x].pb(y);
g[y].pb(x);
}
dfs1(rt,0);
dfs2(rt,rt);
build(1,1,n);
while(q--){
int op,x,y,z;
scanf("%d",&op);
if(op==1){//把x节点到y节点的路径所有点加z
scanf("%d%d%d",&x,&y,&z);
up1(x,y,z);
}
else if(op==2){//查询x节点到y节点的路径和
scanf("%d%d",&x,&y);
qu1(x,y);
}
else if(op==3){////把x节点的所有子孙加y
scanf("%d%d",&x,&y);
up(1,1,n,id[x],id[x]+siz[x]-1,y);
}
else{//查询x节点的子孙和
scanf("%d",&x);
printf("%d\n",qu(1,1,n,id[x],id[x]+siz[x]-1)%mod);
}
}
}
LCA
#define dbg(x) cout<<#x<<" = "<< (x)<< endl
#define IO std::ios::sync_with_stdio(0);
#include <bits/stdc++.h>
#define iter ::iterator
using namespace std;
typedef long long ll;
typedef pair<ll,ll>P;
#define pb push_back
#define se second
#define fi first
#define rs o<<1|1
#define ls o<<1
const ll inf=0x7fffffff;
const int N=5e5+10;
int n,m,s;
vector<int>g[N];
int d[N],lg[N],p[N][22];
void dfs(int u,int fa){
p[u][0]=fa;
d[u]=d[fa]+1;
for(int i=1;(1<<i)<=d[u];i++){
p[u][i]=p[p[u][i-1]][i-1];
}
for(int i=0;i<g[u].size();i++){
int v=g[u][i];
if(v==fa)continue;
dfs(v,u);
}
}
void init(){
for(int i=1;i<=n;i++){
lg[i]=lg[i-1]+((1<<lg[i-1])==i);
}
}
int lca(int x,int y){
if(d[x]<d[y])swap(x,y);
while(d[x]>d[y]){
x=p[x][lg[d[x]-d[y]]-1];
}
if(x==y)return x;
for(int i=lg[x];i>=0;i--){
if(p[x][i]!=p[y][i]){
x=p[x][i];
y=p[y][i];
}
}
return p[x][0];
}
int main(){
//IO;
cin>>n>>m>>s;
for(int i=1;i<n;i++){
int x,y;
scanf("%d%d",&x,&y);
g[x].pb(y);
g[y].pb(x);
}
init();
dfs(s,0);
while(m--){
int x,y;
scanf("%d%d",&x,&y);
//cout<<lca(x,y)<<endl;
printf("%d\n",lca(x,y));
}
}
并查集、倍增
#include <cmath>
#include <cstdio>
#include <iostream>
using namespace std;
const int maxn = 100005, mod = 1000000007;
int n, m, fa[maxn][18], ans;
int find(int x, int k) {
return fa[x][k] == x ? x : fa[x][k] = find(fa[x][k], k);
}
void merge(int x, int y, int k) {
x = find(x, k), y = find(y, k);
if(x != y) fa[x][k] = y;
}
int main() {
scanf("%d %d", &n, &m);
const int maxk = floor(log2(n));
for(int i = 1; i <= n; ++i)
for(int k = 0; k <= maxk; ++k)
fa[i][k] = i;
for(int i = 1, l1, r1, l2, r2; i <= m; ++i) {
scanf("%d %d %d %d", &l1, &r1, &l2, &r2);
for(int k = maxk; ~k; --k)
if(l1+(1<<k)-1 <= r1) merge(l1, l2, k), l1 += 1<<k, l2 += 1<<k;
}
for(int k = maxk; k; --k)
for(int i = 1; i+(1<<k)-1 <= n; ++i) {
int pos = find(i, k);
merge(i, pos, k-1), merge(i+(1<<k-1), pos+(1<<k-1), k-1);
}
for(int i = 1; i <= n; ++i)
if(fa[i][0] == i) ans = !ans ? 9 : ans * 10ll % mod;
printf("%d\n", ans);
return 0;
}
笛卡尔树
#define IO std::ios::sync_with_stdio(0),cin.tie(0),cout.tie(0)
#define bug(x) cout<<#x<<" is "<<x<<endl
#include <bits/stdc++.h>
#define iter ::iterator
using namespace std;
typedef long long ll;
const int N=1e5+5;
int n;
ll a[N],b[N];
int st[N],ls[N],rs[N];
ll ans;
ll dfs(int u){
if(!u)return 0;
ll res=dfs(ls[u])+dfs(rs[u])+b[u];
ll res1=min(res,a[u]);
res1*=res1;
ans=max(ans,res1);
return res;
}
int main(){
IO;
cin>>n;
for(int i=1;i<=n;i++)cin>>a[i]>>b[i];
int h=0;
for(int i=1;i<=n;i++){
int k=h;
while(k>0&&a[st[k]]>a[i])k--;
if(k>0)rs[st[k]]=i;
if(k<h) ls[i]=st[k+1];
st[++k]=i;
h=k;
}
ans=0;
dfs(st[1]);
cout<<ans<<endl;
}
/*
11
9 3 7 1 8 12 10 20 15 18 5
*/
JAVA大数
import java.math.RoundingMode;
import java.math.BigDecimal;
import java.math.BigInteger;
import java.util.Scanner;
import java.util.*;
import java.io.*;
public class Main {
public static long mod=(long)(1e9+7);//1e9会被java看作double
public static long fastpow(long x, long y) {//快速幂
long res=1;
while(y>0) {
if((y&1)==1) {
res=(res*x)%mod;
}
x=(x*x)%mod;
y>>=1;
}
return res;
}
public static int N=200005;
public static long[] b=new long[N];//声明数组
public static void init(int n){//求n的阶乘
b[0]=1;
for(int i=1;i<=n;i++){
b[i]=(b[i-1]*i)%mod;
}
}
public static void main(String [] args){
Scanner cin = new Scanner(System.in);
int T = cin.nextInt();
while (T-- > 0) {
BigInteger x = cin.nextBigInteger();//输入大数
BigInteger y = cin.nextBigInteger();
BigInteger z = BigInteger.valueOf(123456789L);//L标记是long类型
BigInteger k = BigInteger.ZERO;//赋值k为大数类型的0
System.out.println(k);
System.out.println(fastpow(2L,x.longValue()));//快速幂输出2的x次方
if(x.compareTo(y)==0)System.out.println("x=y");//大数的比较函数
else if(x.compareTo(y)<0)System.out.println("x<y");
else if(x.compareTo(y)>0)System.out.println("x>y");
String s=x.toString(2);//把十进制大数转化为二进制数存入字符串。
int n=s.length();//获取字符串长度
for (int i = 0; i < n; i++) {
System.out.print(s.charAt(i) + " ");//空格隔开输出字符.
}
System.out.println();
BigInteger sum = x.add(y);
BigInteger difference = x.subtract(y);
BigInteger product = x.multiply(y);
BigInteger quotient = x.divide(y); // 假设y不为0
BigInteger remainder = x.remainder(y); // 假设y不为0
BigInteger gcd = x.gcd(y); // 最大公约数
BigInteger modInverse = (y.signum() > 0 && x.gcd(y).equals(BigInteger.ONE)) ? x.modInverse(y) : null; //求逆元
BigInteger power = x.pow(2); // x的2次幂
BigInteger shiftedLeft = x.shiftLeft(1); // x左移1位
BigInteger shiftedRight = x.shiftRight(1); // x右移1位
BigInteger andResult = x.and(y); // 位与
BigInteger orResult = x.or(y); // 位或
BigInteger xorResult = x.xor(y); // 位异或
System.out.println("x + y = " + sum);
System.out.println("x - y = " + difference);
System.out.println("x * y = " + product);
System.out.println("x / y = " + quotient);
System.out.println("x % y = " + remainder);
System.out.println("gcd(x, y) = " + gcd);
System.out.println("x modInverse y = " + (modInverse != null ? modInverse : "undefined (no inverse)"));
System.out.println("x^2 = " + power);
System.out.println("x << 1 = " + shiftedLeft);
System.out.println("x >> 1 = " + shiftedRight);
System.out.println("x & y = " + andResult);
System.out.println("x | y = " + orResult);
System.out.println("x ^ y = " + xorResult);
BigDecimal x1 = new BigDecimal(cin.next()); // 读取高精度小数
BigDecimal y1 = new BigDecimal(cin.next());
BigDecimal sum1 = x1.add(y1);
BigDecimal difference1 = x1.subtract(y1);
BigDecimal product1 = x1.multiply(y1);
BigDecimal quotient1 = x1.divide(y1, 10, RoundingMode.HALF_UP); // 设置结果精度为10位小数并使用四舍五入
System.out.println("x + y = " + sum1);
System.out.println("x - y = " + difference1);
System.out.println("x * y = " + product1);
System.out.println("x / y = " + quotient1); // 假设y不为0
BigDecimal original = new BigDecimal("123.456789");
BigDecimal result1 = original.setScale(3, RoundingMode.HALF_UP);// 小数点后保留3位，采用四舍五入
System.out.println(result1); // 输出：123.457
BigDecimal dividend1 = new BigDecimal("10");
BigDecimal divisor1 = new BigDecimal("3");
// 指定精度为10位小数，使用四舍五入模式
result1 = dividend1.divide(divisor1, 10, RoundingMode.HALF_UP);
System.out.println(result1); // 输出: 3.3333333333
}
cin.close();
}
}
/*
c=a.max(b); // 取a，b中较大的，a.min(b)就是取较小的
d=a.compareTo(b); //比较a与b的大小 d=-1小于d=0等于 d=1大于 d为int型
a.equals(b); // 判断a与b是否相等 相等返回true 不相等返回false
d=a.intValue(); // 将大数a转换为 int 类型赋值给 d
e=a.longValue(); // 将大数a转换为 long 类型赋值给 e
f=a.floatValue(); // 将大数a转换为 float 类型赋值给 f
g=a.doubleValue(); // 将大数a转换为 double 类型赋值给 g
s=a.toString(); // 将大数a转换为 String 类型赋值给 s
s=a.toPlainString(); //将大数a转换为String类型赋值给s，且不表示为科学计数法
a=BigInteger.valueOf(e); // 将 e 以大数形式赋值给大数 a e只能为long或int
a=newBigInteger(s, d); // 将s数字字符串以d进制赋值给大数a如果d=s字符数字的进制则等同于将数字字符串以大数形式赋值给大数a
String st = Integer.toString(num, base); //把int型num当10进制的数转成base进制数存入st中(base <= 35).
int num = Integer.parseInt(st, base); //把st当做base进制，转成10进制的int
(parseInt有两个参数,第一个为要转的字符串,第二个为说明是什么进制).
BigInter m = new BigInteger(st, base); // st是字符串，base是st的进制.
a=cin.nextBigInteger(b); //以b进制读入一个大数赋值给a
c=a.toString(b); // 将大数a以b进制的方式赋给字符串c
a=newBigInteger(c, b); //把c 当做“b进制“转为十进制大数赋值给a
*/
计算几何
链接：https://ac.nowcoder.com/acm/contest/1112/J
来源：牛客网
Bobo 有一个三角形和一个矩形，他想求他们交的面积。
具体地，三角形和矩形由 8 个整数 x1,y1,x2,y2,x3,y3,x4,y4x_1, y_1, x_2, y_2, x_3, y_3, x_4, y_4x1,y1,x2,y2,x3,y3,x4,y4
描述。
表示三角形的顶点坐标是 (x1,y1),(x1,y2),(x2,y1)(x_1,
y_1), (x_1, y_2), (x_2, y_1)(x1,y1),(x1,y2),(x2,y1)，
矩形的顶点坐标是 (x3,y3),(x3,y4),(x4,y4),(x4,y3)(x_3,
y_3), (x_3, y_4), (x_4, y_4), (x_4, y_3)(x3,y3),(x3,y4),(x4,y4),(x4,y3)
把三角形的顶点里在矩形里面的点放进数组；
把矩形的顶点里在三角形里面的点放进数组；
把三角形三条边和矩形四条边的交点放进数组（规范相交）；
对这个数组去重并求凸包然后求凸包面积就是答案。
#include <bits/stdc++.h>
using namespace std;
#define ll long long
const int N=1e3+10;
double eps=1e-8;
double pi=acos(-1);
struct Point{
double x,y;
Point(double x=0,double y=0):x(x),y(y){};
};
typedef Point Vector;
Vector operator +(Vector A,Vector B){return Vector(A.x+B.x,A.y+B.y);}
Vector operator -(Vector A,Vector B){return Vector(A.x-B.x,A.y-B.y);}
Vector operator *(Vector A,double B){return Vector(A.x*B,A.y*B);}
Vector operator /(Vector A,double B){return Vector(A.x/B,A.y/B);}
int dcmp(double x){
if(fabs(x)<eps)return 0;
return x<0?-1:1;
}
bool operator<(const Point&a,const Point&b){return a.x<b.x||(a.x==b.x&&a.y<b.y);}
bool operator == (const Point &a,const Point &b){return dcmp(a.x-b.x)==0&&dcmp(a.y-b.y)==0;}
double Dot(Vector A,Vector B){return A.x*B.x+A.y*B.y;}
double Length(Vector A){return sqrt(Dot(A,A));}
double Cross(Vector A,Vector B){return A.x*B.y-A.y*B.x;}
double Angle(Vector A,Vector B){return acos(Dot(A,B)/Length(A)/Length(B));}
double Angle(Vector A) {return atan2(A.y, A.x);}
struct Circle{ //圆
Point c;
double r;
Circle(Point c = Point(0, 0), double r = 0):c(c),r(r){}
Point point(double a){
return Point(c.x+cos(a)*r,c.y+sin(a)*r);
}
};
double dis(Point A,Point B){
return sqrt((A.x-B.x)*(A.x-B.x)+(A.y-B.y)*(A.y-B.y));
}
int circle_circle(Circle C1,Circle C2,vector<Point>&sol){//返回圆与圆的交点个数
double d=Length(C1.c-C2.c);
if(dcmp(d)==0){
if(dcmp(C1.r-C2.r)==0)return -1;
return 0;
}
if(dcmp(C1.r+C2.r-d)<0)return 0;
if(dcmp(fabs(C1.r-C2.r)-d)>0)return 0;
double a=Angle(C2.c-C1.c);
double da=acos((C1.r*C1.r+d*d-C2.r*C2.r)/(2*d*C1.r));
Point p1=C1.point(a-da),p2=C1.point(a+da);
sol.push_back(p1);
if(p1==p2)return 1;
sol.push_back(p2);
return 2;
}
Point line_point(Point p,Vector v,Point q,Vector w){//直线交点
Vector u=p-q;
double t=Cross(w,u)/Cross(v,w);
return p+v*t;
}
bool onsegment(Point p,Point a1,Point a2){
if(p==a1||p==a2)return true;
return dcmp(Cross(a1-p,a2-p))==0&&dcmp(Dot(a1-p,a2-p))<0;
}
bool segmentcross(Point a1,Point a2,Point b1,Point b2){
double c1=Cross(a2-a1,b1-a1),c2=Cross(a2-a1,b2-a1),
c3=Cross(b2-b1,a1-b1),c4=Cross(b2-b1,a2-b1);
return dcmp(c1)*dcmp(c2)<0&&dcmp(c3)*dcmp(c4)<0;
}
Point line_line(Point A,Point B,Point C,Vector D){//线段交点
if(segmentcross(A,B,C,D)){
return line_point(A,B-A,C,D-C);
}
return Point(123.456,0);
}
int tubao(Point *p,int n,Point *ch){
sort(p,p+n);
n=unique(p,p+n)-p;
int m=0;
for(int i=0;i<n;i++){
while(m>1&&Cross(ch[m-1]-ch[m-2],p[i]-ch[m-2])<=0)m--;
ch[m++]=p[i];
}
int k=m;
for(int i=n-2;i>=0;i--){
while(m>k&&Cross(ch[m-1]-ch[m-2],p[i]-ch[m-2])<=0)m--;
ch[m++]=p[i];
}
if(n>1)m--;
return m;
}
double tubaos(Point *p,int n){
double area=0;
for(int i=1;i<n-1;i++){
area+=Cross(p[i]-p[0],p[i+1]-p[0]);
}
return area/2;
}
int intubao(Point *ch,int n,Point p){
Vector A,B;
int flag=0;
for(int i=0;i<n;i++){
A=ch[(i+1)%n]-ch[i];
B=p-ch[i];
if(onsegment(p,ch[i],ch[(i+1)%n])){
flag=-1;
break;
}
else if(Cross(A,B)>0){
flag++;
}
}
if(flag==-1||flag==n)return 1;
return 0;
}
double x[N],y[N];
Point p[N],q[N],ch[N],ans[N];
int main(){
while(~scanf("%lf%lf",&x[1],&y[1])){
scanf("%lf%lf",&x[2],&y[2]);
scanf("%lf%lf",&x[3],&y[3]);
scanf("%lf%lf",&x[4],&y[4]);
int n=0;
p[n++]=Point(x[3],y[3]);
p[n++]=Point(x[4],y[3]);
p[n++]=Point(x[4],y[4]);
p[n++]=Point(x[3],y[4]);
int m=0;
q[m++]=Point(x[1],y[1]);
q[m++]=Point(x[1],y[2]);
q[m++]=Point(x[2],y[1]);
tubao(q,m,ch);
int cnt=0;
for(int i=0;i<n;i++){
if(intubao(ch,m,p[i])){
ans[cnt++]=p[i];
}
}
for(int i=0;i<m;i++){
if(intubao(p,n,q[i])){
ans[cnt++]=q[i];
}
}
p[n]=p[0];
q[m]=q[0];
for(int i=0;i<n;i++){
for(int j=0;j<m;j++){
Point A=line_line(p[i],p[i+1],q[j],q[j+1]);
if(dcmp(A.x-123.456)!=0)ans[cnt++]=A;
}
}
int n1=tubao(ans,cnt,ch);
if(n1<=2)printf("%.8lf\n",0.0);
else printf("%.8lf\n",tubaos(ch,n1));
}
}
字符串双哈希
https://codeforc.es/contest/1200/problem/E
题意：给定一堆字符串，依次插入答案串尾部，每次删掉答案串的后缀 与 待插入串的前缀的最大匹配串
思路：hash匹配后缀和前缀，不断更新hash值
#define bug(x) cout<<#x<<" is "<<x<<endl
#define IO std::ios::sync_with_stdio(0)
#include <bits/stdc++.h>
#define iter ::iterator
#define pa pair<int,int>
using namespace std;
#define ll long long
#define mk make_pair
#define pb push_back
#define se second
#define fi first
#define ls o<<1
#define rs o<<1|1
const int N=1e6+5;
int n;
ll mod[5]={1000000009,998244353};
ll base[5]={666127,777131};
ll p[N][2];
ll h[N][2];
void init(){
for(int i=0;i<2;i++)p[0][i]=1;
for(int i=1;i<=N-2;i++){
for(int j=0;j<2;j++){
p[i][j]=p[i-1][j]*base[j]%mod[j];
}
}
}
ll cal(int l,int r,int t){
return (h[r][t]-h[l-1][t]*p[(r-l+1)][t]%mod[t]+mod[t])%mod[t];
}
ll get(char c){
if('a'<=c&&c<='z')return c-'a';
if('A'<=c&&c<='Z')return c-'A'+26;
if('0'<=c&&c<='9')return c-'0'+52;
}
string s,t;
int main(){
IO;
cin>>n;
cin>>s;
int pre=1;
init();
for(int g=1;g<n;g++){
cin>>t;
int ns=s.length();
for(int i=pre;i<=ns;i++){
ll c=get(s[i-1]);
for(int j=0;j<2;j++){
h[i][j]=(h[i-1][j]*base[j]%mod[j]+c)%mod[j];
}
}
int nt=t.length();
ll H[2]={0};
pre=ns+1;
int id=0;
for(int i=1;i<=nt;i++){
int f=0;
ll c=get(t[i-1]);
for(int j=0;j<2;j++){
H[j]=(H[j]*base[j]%mod[j]+c)%mod[j];
if(cal(ns-i+1,ns,j)==H[j])f++;
}
if(f==2){
id=i;
}
if(i==ns)break;
}
s+=t.substr(id,nt-id);
}
cout<<s<<endl;
}
/*
7
RAtONC14epz KfIenADgDKDci OMmOOQOc yVrfGLV49fW1 xntodZLM5 2f7LXdzX xIhm
5
I want to order pizza
*/
离散化和部分常用STL操作
sort(a,a+n);
int m=unique(a,a+n)-a;
for(int i=0;i<n;i++) b[i]=lower_bound(a,a+m,b[i])-a+1;
lower_bound(v.begin(), v.end(), x)//vector的lower_bound用法
multiset<int>s;
s.insert(x);
s.erase(x)//既可以删除与x相等的所有元素，也可以删除单个迭代器
s.erase(L,R)//删除迭代器[L,R)内的所有元素，是左闭右开区间。
auto id=s.lower_lound(x);
auto id=s.upper_bound(x);
List双向链表
#include <bits/stdc++.h>
using namespace std;
int main() {
list<int> list1 = {1, 2, 3, 4, 5}; // 创建两个 list
list<int> list2 = {10, 20, 30, 40, 50};
list1.splice(list1.begin(), list2); // 1. 使用 splice 移动整个 list2 到 list1 的开始位置
//list1: 10 20 30 40 50 1 2 3 4 5
//list2: 为空
auto it=find(list1.begin(),list2.end(),3);//指定list1某个位置移动到list1最前面
list1.splice(list1.begin(),list1,it);
//3 10 20 30 40 50 1 2 4 5
// 3. 使用 splice 移动 list1 中的一段元素 [20, 40] 到 list2 的结束位置
auto first = find(list1.begin(), list1.end(), 20);
auto last = find(list1.begin(), list1.end(), 40);
if (first != list1.end() && last != list1.end()) {
last = next(last); // 使 last往后移动一次，再将[fisrt,last)区间内的数移到指定位置
list2.splice(list2.end(), list1, first, last);//将list1的[first,last]中元素移动到list2末尾
}
// list1: 3 10 50 1 2 4 5
// list2: 20 30 40
it = find(list1.begin(), list1.end(), 50);
auto it1=next(it); // 得到it的下一个迭代器，指向的数为1
auto it2=prev(it); // 得到it的上一个迭代器，指向的数为10
cout<<*it1<<" "<<*it2<<endl;
}
主席树区间修改
#define IO std::ios::sync_with_stdio(0),cin.tie(0),cout.tie(0)
#define bug(x) cout<<#x<<" is "<<x<<endl
#include <bits/stdc++.h>
#define iter ::iterator
using namespace std;
typedef long long ll;
typedef pair<int,int>P;
#define pb push_back
#define mk make_pair
#define se second
#define fi first
const ll mod=1e9+7;
const int N=1e5+5;
int n,q;
int a[N],rt[N*25],ls[N*25],rs[N*25];
ll sum[N*25],add[N*25];
int cnt;
void build(int &o,int l,int r){
o=++cnt;
if(l==r){
sum[o]=a[l];
return;
}
int m=(l+r)/2;
build(ls[o],l,m);
build(rs[o],m+1,r);
sum[o]=sum[ls[o]]+sum[rs[o]];
}
void up(int &o,int pre,int l,int r,int ql,int qr,int val){
o=++cnt;
ls[o]=ls[pre];
rs[o]=rs[pre];
add[o]=add[pre];
sum[o]=sum[pre]+1ll*val*(min(r,qr)-max(l,ql)+1);
if(l>=ql&&r<=qr){
add[o]+=val;
return;
}
int m=(l+r)/2;
if(ql<=m)up(ls[o],ls[pre],l,m,ql,qr,val);
if(qr>m)up(rs[o],rs[pre],m+1,r,ql,qr,val);
}
ll qu(int o,int l,int r,int ql,int qr){
if(l>=ql&&r<=qr)return sum[o];
ll res=add[o]*(min(r,qr)-max(l,ql)+1);
int m=(l+r)/2;
if(ql<=m)res+=qu(ls[o],l,m,ql,qr);
if(qr>m)res+=qu(rs[o],m+1,r,ql,qr);
return res;
}
int main(){
while(~scanf("%d%d",&n,&q)){
for(int i=1;i<=n;i++)scanf("%d",&a[i]);
cnt=0;
build(rt[0],1,n);
int time=0;
while(q--){
char op;
int L,R,val;
scanf(" %c",&op);
if(op=='Q'){
scanf("%d%d",&L,&R);
ll ans=qu(rt[time],1,n,L,R);
printf("%lld\n",ans);
}
else if(op=='C'){
scanf("%d%d%d",&L,&R,&val);
++time;
up(rt[time],rt[time-1],1,n,L,R,val);
}
else if(op=='H'){
scanf("%d%d%d",&L,&R,&val);
ll ans=qu(rt[val],1,n,L,R);
printf("%lld\n",ans);
}
else{
scanf("%d",&val);
time=val;
}
}
printf("\n");
}
}
带权并查集
#include <bits/stdc++.h>
using namespace std;
const int N = 200010;
int parent[N];
int value[N]; ///记录当前结点到根结点的距离
int FindSet(int x) {
if (x == parent[x])
return x;
else {
int t = parent[x]; //记录原父节点编号
parent[x] = FindSet(parent[x]); //父节点变为根节点，此时value[x]=父节点到根节点的权值
value[x] += value[t]; //当前节点的权值加上原本父节点的权值
return parent[x];
}
}
int main(){
int n, m;
while (scanf("%d%d", &n, &m) != EOF) {
for (int i = 0; i <= n; i++) {
parent[i] = i;
value[i] = 0;
}
int ans = 0;
while (m--) {
int a, b, v;
scanf("%d%d%d", &a, &b, &v);
a--;
int roota = FindSet(a);
int rootb = FindSet(b);
if (roota == rootb) {
if (value[a] - value[b] != v) ans++; ///精华部分1
}
else {
parent[roota] = rootb;
value[roota] = -value[a] + value[b] + v; ///精华部分2
}
}
printf("%d\n", ans);
}
}
树剖
#define IO std::ios::sync_with_stdio(0);
#include <bits/stdc++.h>
#define iter ::iterator
using namespace std;
typedef long long ll;
typedef pair<ll,ll>P;
#define pb push_back
#define se second
#define fi first
#define rs o<<1|1
#define ls o<<1
#define inf 0x3f3f3f3f
const int N=1e5+5;
int n,q,rt,mod;
int a[N],b[N];
vector<int>g[N];
int deep[N],fa[N],id[N],son[N],siz[N],top[N];
int sumv[N*4],add[N*4];
int L[N*4],R[N*4];
void dfs1(int u,int f){
deep[u]=deep[f]+1;
fa[u]=f;
siz[u]=1;
int mx=-1;
for(int i=0;i<g[u].size();i++){
int v=g[u][i];
if(v==f)continue;
dfs1(v,u);
siz[u]+=siz[v];
if(siz[v]>mx){
mx=siz[v];
son[u]=v;
}
}
}
int tim;
void dfs2(int u,int topf){
id[u]=++tim;
a[tim]=b[u];
top[u]=topf;
if(!son[u])return;
dfs2(son[u],topf);
for(int i=0;i<g[u].size();i++){
int v=g[u][i];
if(v==fa[u]||v==son[u])continue;
dfs2(v,v);
}
}
void push(int o){
sumv[o]=(sumv[ls]+sumv[rs])%mod;
}
void down(int o){
if(add[o]){
add[o]%=mod;
sumv[ls]+=add[o]*(R[ls]-L[ls]+1);
sumv[ls]%=mod;
sumv[rs]+=add[o]*(R[rs]-L[rs]+1);
sumv[rs]%=mod;
add[ls]+=add[o];
add[ls]%=mod;
add[rs]+=add[o];
add[rs]%=mod;
add[o]=0;
}
}
void build(int o,int l,int r){
if(l==r){
sumv[o]=a[l]%mod;
L[o]=R[o]=l;
return;
}
int m=(l+r)/2;
build(ls,l,m);
build(rs,m+1,r);
L[o]=L[ls];
R[o]=R[rs];
push(o);
}
int qu(int o,int l,int r,int ql,int qr){
if(l>=ql&&r<=qr){
return sumv[o]%mod;
}
down(o);
int m=(l+r)/2;
int res=0;
if(ql<=m)res+=qu(ls,l,m,ql,qr)%mod;
res%=mod;
if(qr>m)res+=qu(rs,m+1,r,ql,qr)%mod;
res%=mod;
return res;
}
void up(int o,int l,int r,int ql,int qr,int v){
if(l>=ql&&r<=qr){
v%=mod;
add[o]+=v;
add[o]%=mod;
sumv[o]+=v*(r-l+1);
sumv[o]%=mod;
return;
}
down(o);
int m=(l+r)/2;
if(ql<=m)up(ls,l,m,ql,qr,v);
if(qr>m)up(rs,m+1,r,ql,qr,v);
push(o);
}
void qu1(int x,int y){//x节点到y节点的路径和
int res=0;
while(top[x]!=top[y]){
if(deep[top[x]]<deep[top[y]])swap(x,y);
res+=qu(1,1,n,id[top[x]],id[x]);
res%=mod;
x=fa[top[x]];
}
if(deep[x]>deep[y])swap(x,y);
res+=qu(1,1,n,id[x],id[y])%mod;
res%=mod;
printf("%d\n",res);
}
void up1(int x,int y,int z){//把x节点到y节点的路径所有点加z
while(top[x]!=top[y]){
if(deep[top[x]]<deep[top[y]])swap(x,y);
up(1,1,n,id[top[x]],id[x],z);
x=fa[top[x]];
}
if(deep[x]>deep[y])swap(x,y);
up(1,1,n,id[x],id[y],z);
}
int main(){
scanf("%d%d%d%d",&n,&q,&rt,&mod);
for(int i=1;i<=n;i++){
scanf("%d",&b[i]);
b[i]%=mod;
}
for(int i=1;i<n;i++){
int x,y;
scanf("%d%d",&x,&y);
g[x].pb(y);
g[y].pb(x);
}
dfs1(rt,0);
dfs2(rt,rt);
build(1,1,n);
while(q--){
int op,x,y,z;
scanf("%d",&op);
if(op==1){//把x节点到y节点的路径所有点加z
scanf("%d%d%d",&x,&y,&z);
up1(x,y,z);
}
else if(op==2){//查询x节点到y节点的路径和
scanf("%d%d",&x,&y);
qu1(x,y);
}
else if(op==3){////把x节点的所有子孙加y
scanf("%d%d",&x,&y);
up(1,1,n,id[x],id[x]+siz[x]-1,y);
}
else{//查询x节点的子孙和
scanf("%d",&x);
printf("%d\n",qu(1,1,n,id[x],id[x]+siz[x]-1)%mod);
}
}
}
LCA
#define dbg(x) cout<<#x<<" = "<< (x)<< endl
#define IO std::ios::sync_with_stdio(0);
#include <bits/stdc++.h>
#define iter ::iterator
using namespace std;
typedef long long ll;
typedef pair<ll,ll>P;
#define pb push_back
#define se second
#define fi first
#define rs o<<1|1
#define ls o<<1
const ll inf=0x7fffffff;
const int N=5e5+10;
int n,m,s;
vector<int>g[N];
int d[N],lg[N],p[N][22];
void dfs(int u,int fa){
p[u][0]=fa;
d[u]=d[fa]+1;
for(int i=1;(1<<i)<=d[u];i++){
p[u][i]=p[p[u][i-1]][i-1];
}
for(int i=0;i<g[u].size();i++){
int v=g[u][i];
if(v==fa)continue;
dfs(v,u);
}
}
void init(){
for(int i=1;i<=n;i++){
lg[i]=lg[i-1]+((1<<lg[i-1])==i);
}
}
int lca(int x,int y){
if(d[x]<d[y])swap(x,y);
while(d[x]>d[y]){
x=p[x][lg[d[x]-d[y]]-1];
}
if(x==y)return x;
for(int i=lg[x];i>=0;i--){
if(p[x][i]!=p[y][i]){
x=p[x][i];
y=p[y][i];
}
}
return p[x][0];
}
int main(){
//IO;
cin>>n>>m>>s;
for(int i=1;i<n;i++){
int x,y;
scanf("%d%d",&x,&y);
g[x].pb(y);
g[y].pb(x);
}
init();
dfs(s,0);
while(m--){
int x,y;
scanf("%d%d",&x,&y);
//cout<<lca(x,y)<<endl;
printf("%d\n",lca(x,y));
}
}
并查集、倍增
#include <cmath>
#include <cstdio>
#include <iostream>
using namespace std;
const int maxn = 100005, mod = 1000000007;
int n, m, fa[maxn][18], ans;
int find(int x, int k) {
return fa[x][k] == x ? x : fa[x][k] = find(fa[x][k], k);
}
void merge(int x, int y, int k) {
x = find(x, k), y = find(y, k);
if(x != y) fa[x][k] = y;
}
int main() {
scanf("%d %d", &n, &m);
const int maxk = floor(log2(n));
for(int i = 1; i <= n; ++i)
for(int k = 0; k <= maxk; ++k)
fa[i][k] = i;
for(int i = 1, l1, r1, l2, r2; i <= m; ++i) {
scanf("%d %d %d %d", &l1, &r1, &l2, &r2);
for(int k = maxk; ~k; --k)
if(l1+(1<<k)-1 <= r1) merge(l1, l2, k), l1 += 1<<k, l2 += 1<<k;
}
for(int k = maxk; k; --k)
for(int i = 1; i+(1<<k)-1 <= n; ++i) {
int pos = find(i, k);
merge(i, pos, k-1), merge(i+(1<<k-1), pos+(1<<k-1), k-1);
}
for(int i = 1; i <= n; ++i)
if(fa[i][0] == i) ans = !ans ? 9 : ans * 10ll % mod;
printf("%d\n", ans);
return 0;
}
笛卡尔树
#define IO std::ios::sync_with_stdio(0),cin.tie(0),cout.tie(0)
#define bug(x) cout<<#x<<" is "<<x<<endl
#include <bits/stdc++.h>
#define iter ::iterator
using namespace std;
typedef long long ll;
const int N=1e5+5;
int n;
ll a[N],b[N];
int st[N],ls[N],rs[N];
ll ans;
ll dfs(int u){
if(!u)return 0;
ll res=dfs(ls[u])+dfs(rs[u])+b[u];
ll res1=min(res,a[u]);
res1*=res1;
ans=max(ans,res1);
return res;
}
int main(){
IO;
cin>>n;
for(int i=1;i<=n;i++)cin>>a[i]>>b[i];
int h=0;
for(int i=1;i<=n;i++){
int k=h;
while(k>0&&a[st[k]]>a[i])k--;
if(k>0)rs[st[k]]=i;
if(k<h) ls[i]=st[k+1];
st[++k]=i;
h=k;
}
ans=0;
dfs(st[1]);
cout<<ans<<endl;
}
/*
11
9 3 7 1 8 12 10 20 15 18 5
*/
JAVA大数
import java.math.RoundingMode;
import java.math.BigDecimal;
import java.math.BigInteger;
import java.util.Scanner;
import java.util.*;
import java.io.*;
public class Main {
public static long mod=(long)(1e9+7);//1e9会被java看作double
public static long fastpow(long x, long y) {//快速幂
long res=1;
while(y>0) {
if((y&1)==1) {
res=(res*x)%mod;
}
x=(x*x)%mod;
y>>=1;
}
return res;
}
public static int N=200005;
public static long[] b=new long[N];//声明数组
public static void init(int n){//求n的阶乘
b[0]=1;
for(int i=1;i<=n;i++){
b[i]=(b[i-1]*i)%mod;
}
}
public static void main(String [] args){
Scanner cin = new Scanner(System.in);
int T = cin.nextInt();
while (T-- > 0) {
BigInteger x = cin.nextBigInteger();//输入大数
BigInteger y = cin.nextBigInteger();
BigInteger z = BigInteger.valueOf(123456789L);//L标记是long类型
BigInteger k = BigInteger.ZERO;//赋值k为大数类型的0
System.out.println(k);
System.out.println(fastpow(2L,x.longValue()));//快速幂输出2的x次方
if(x.compareTo(y)==0)System.out.println("x=y");//大数的比较函数
else if(x.compareTo(y)<0)System.out.println("x<y");
else if(x.compareTo(y)>0)System.out.println("x>y");
String s=x.toString(2);//把十进制大数转化为二进制数存入字符串。
int n=s.length();//获取字符串长度
for (int i = 0; i < n; i++) {
System.out.print(s.charAt(i) + " ");//空格隔开输出字符.
}
System.out.println();
BigInteger sum = x.add(y);
BigInteger difference = x.subtract(y);
BigInteger product = x.multiply(y);
BigInteger quotient = x.divide(y); // 假设y不为0
BigInteger remainder = x.remainder(y); // 假设y不为0
BigInteger gcd = x.gcd(y); // 最大公约数
BigInteger modInverse = (y.signum() > 0 && x.gcd(y).equals(BigInteger.ONE)) ? x.modInverse(y) : null; //求逆元
BigInteger power = x.pow(2); // x的2次幂
BigInteger shiftedLeft = x.shiftLeft(1); // x左移1位
BigInteger shiftedRight = x.shiftRight(1); // x右移1位
BigInteger andResult = x.and(y); // 位与
BigInteger orResult = x.or(y); // 位或
BigInteger xorResult = x.xor(y); // 位异或
System.out.println("x + y = " + sum);
System.out.println("x - y = " + difference);
System.out.println("x * y = " + product);
System.out.println("x / y = " + quotient);
System.out.println("x % y = " + remainder);
System.out.println("gcd(x, y) = " + gcd);
System.out.println("x modInverse y = " + (modInverse != null ? modInverse : "undefined (no inverse)"));
System.out.println("x^2 = " + power);
System.out.println("x << 1 = " + shiftedLeft);
System.out.println("x >> 1 = " + shiftedRight);
System.out.println("x & y = " + andResult);
System.out.println("x | y = " + orResult);
System.out.println("x ^ y = " + xorResult);
BigDecimal x1 = new BigDecimal(cin.next()); // 读取高精度小数
BigDecimal y1 = new BigDecimal(cin.next());
BigDecimal sum1 = x1.add(y1);
BigDecimal difference1 = x1.subtract(y1);
BigDecimal product1 = x1.multiply(y1);
BigDecimal quotient1 = x1.divide(y1, 10, RoundingMode.HALF_UP); // 设置结果精度为10位小数并使用四舍五入
System.out.println("x + y = " + sum1);
System.out.println("x - y = " + difference1);
System.out.println("x * y = " + product1);
System.out.println("x / y = " + quotient1); // 假设y不为0
BigDecimal original = new BigDecimal("123.456789");
BigDecimal result1 = original.setScale(3, RoundingMode.HALF_UP);// 小数点后保留3位，采用四舍五入
System.out.println(result1); // 输出：123.457
BigDecimal dividend1 = new BigDecimal("10");
BigDecimal divisor1 = new BigDecimal("3");
// 指定精度为10位小数，使用四舍五入模式
result1 = dividend1.divide(divisor1, 10, RoundingMode.HALF_UP);
System.out.println(result1); // 输出: 3.3333333333
}
cin.close();
}
}
/*
c=a.max(b); // 取a，b中较大的，a.min(b)就是取较小的
d=a.compareTo(b); //比较a与b的大小 d=-1小于d=0等于 d=1大于 d为int型
a.equals(b); // 判断a与b是否相等 相等返回true 不相等返回false
d=a.intValue(); // 将大数a转换为 int 类型赋值给 d
e=a.longValue(); // 将大数a转换为 long 类型赋值给 e
f=a.floatValue(); // 将大数a转换为 float 类型赋值给 f
g=a.doubleValue(); // 将大数a转换为 double 类型赋值给 g
s=a.toString(); // 将大数a转换为 String 类型赋值给 s
s=a.toPlainString(); //将大数a转换为String类型赋值给s，且不表示为科学计数法
a=BigInteger.valueOf(e); // 将 e 以大数形式赋值给大数 a e只能为long或int
a=newBigInteger(s, d); // 将s数字字符串以d进制赋值给大数a如果d=s字符数字的进制则等同于将数字字符串以大数形式赋值给大数a
String st = Integer.toString(num, base); //把int型num当10进制的数转成base进制数存入st中(base <= 35).
int num = Integer.parseInt(st, base); //把st当做base进制，转成10进制的int
(parseInt有两个参数,第一个为要转的字符串,第二个为说明是什么进制).
BigInter m = new BigInteger(st, base); // st是字符串，base是st的进制.
a=cin.nextBigInteger(b); //以b进制读入一个大数赋值给a
c=a.toString(b); // 将大数a以b进制的方式赋给字符串c
a=newBigInteger(c, b); //把c 当做“b进制“转为十进制大数赋值给a
*/
计算几何
链接：https://ac.nowcoder.com/acm/contest/1112/J
来源：牛客网
Bobo 有一个三角形和一个矩形，他想求他们交的面积。
具体地，三角形和矩形由 8 个整数 x1,y1,x2,y2,x3,y3,x4,y4x_1, y_1, x_2, y_2, x_3, y_3, x_4, y_4x1,y1,x2,y2,x3,y3,x4,y4
描述。
表示三角形的顶点坐标是 (x1,y1),(x1,y2),(x2,y1)(x_1,
y_1), (x_1, y_2), (x_2, y_1)(x1,y1),(x1,y2),(x2,y1)，
矩形的顶点坐标是 (x3,y3),(x3,y4),(x4,y4),(x4,y3)(x_3,
y_3), (x_3, y_4), (x_4, y_4), (x_4, y_3)(x3,y3),(x3,y4),(x4,y4),(x4,y3)
把三角形的顶点里在矩形里面的点放进数组；
把矩形的顶点里在三角形里面的点放进数组；
把三角形三条边和矩形四条边的交点放进数组（规范相交）；
对这个数组去重并求凸包然后求凸包面积就是答案。
#include <bits/stdc++.h>
using namespace std;
#define ll long long
const int N=1e3+10;
double eps=1e-8;
double pi=acos(-1);
struct Point{
double x,y;
Point(double x=0,double y=0):x(x),y(y){};
};
typedef Point Vector;
Vector operator +(Vector A,Vector B){return Vector(A.x+B.x,A.y+B.y);}
Vector operator -(Vector A,Vector B){return Vector(A.x-B.x,A.y-B.y);}
Vector operator *(Vector A,double B){return Vector(A.x*B,A.y*B);}
Vector operator /(Vector A,double B){return Vector(A.x/B,A.y/B);}
int dcmp(double x){
if(fabs(x)<eps)return 0;
return x<0?-1:1;
}
bool operator<(const Point&a,const Point&b){return a.x<b.x||(a.x==b.x&&a.y<b.y);}
bool operator == (const Point &a,const Point &b){return dcmp(a.x-b.x)==0&&dcmp(a.y-b.y)==0;}
double Dot(Vector A,Vector B){return A.x*B.x+A.y*B.y;}
double Length(Vector A){return sqrt(Dot(A,A));}
double Cross(Vector A,Vector B){return A.x*B.y-A.y*B.x;}
double Angle(Vector A,Vector B){return acos(Dot(A,B)/Length(A)/Length(B));}
double Angle(Vector A) {return atan2(A.y, A.x);}
struct Circle{ //圆
Point c;
double r;
Circle(Point c = Point(0, 0), double r = 0):c(c),r(r){}
Point point(double a){
return Point(c.x+cos(a)*r,c.y+sin(a)*r);
}
};
double dis(Point A,Point B){
return sqrt((A.x-B.x)*(A.x-B.x)+(A.y-B.y)*(A.y-B.y));
}
int circle_circle(Circle C1,Circle C2,vector<Point>&sol){//返回圆与圆的交点个数
double d=Length(C1.c-C2.c);
if(dcmp(d)==0){
if(dcmp(C1.r-C2.r)==0)return -1;
return 0;
}
if(dcmp(C1.r+C2.r-d)<0)return 0;
if(dcmp(fabs(C1.r-C2.r)-d)>0)return 0;
double a=Angle(C2.c-C1.c);
double da=acos((C1.r*C1.r+d*d-C2.r*C2.r)/(2*d*C1.r));
Point p1=C1.point(a-da),p2=C1.point(a+da);
sol.push_back(p1);
if(p1==p2)return 1;
sol.push_back(p2);
return 2;
}
Point line_point(Point p,Vector v,Point q,Vector w){//直线交点
Vector u=p-q;
double t=Cross(w,u)/Cross(v,w);
return p+v*t;
}
bool onsegment(Point p,Point a1,Point a2){
if(p==a1||p==a2)return true;
return dcmp(Cross(a1-p,a2-p))==0&&dcmp(Dot(a1-p,a2-p))<0;
}
bool segmentcross(Point a1,Point a2,Point b1,Point b2){
double c1=Cross(a2-a1,b1-a1),c2=Cross(a2-a1,b2-a1),
c3=Cross(b2-b1,a1-b1),c4=Cross(b2-b1,a2-b1);
return dcmp(c1)*dcmp(c2)<0&&dcmp(c3)*dcmp(c4)<0;
}
Point line_line(Point A,Point B,Point C,Vector D){//线段交点
if(segmentcross(A,B,C,D)){
return line_point(A,B-A,C,D-C);
}
return Point(123.456,0);
}
int tubao(Point *p,int n,Point *ch){
sort(p,p+n);
n=unique(p,p+n)-p;
int m=0;
for(int i=0;i<n;i++){
while(m>1&&Cross(ch[m-1]-ch[m-2],p[i]-ch[m-2])<=0)m--;
ch[m++]=p[i];
}
int k=m;
for(int i=n-2;i>=0;i--){
while(m>k&&Cross(ch[m-1]-ch[m-2],p[i]-ch[m-2])<=0)m--;
ch[m++]=p[i];
}
if(n>1)m--;
return m;
}
double tubaos(Point *p,int n){
double area=0;
for(int i=1;i<n-1;i++){
area+=Cross(p[i]-p[0],p[i+1]-p[0]);
}
return area/2;
}
int intubao(Point *ch,int n,Point p){
Vector A,B;
int flag=0;
for(int i=0;i<n;i++){
A=ch[(i+1)%n]-ch[i];
B=p-ch[i];
if(onsegment(p,ch[i],ch[(i+1)%n])){
flag=-1;
break;
}
else if(Cross(A,B)>0){
flag++;
}
}
if(flag==-1||flag==n)return 1;
return 0;
}
double x[N],y[N];
Point p[N],q[N],ch[N],ans[N];
int main(){
while(~scanf("%lf%lf",&x[1],&y[1])){
scanf("%lf%lf",&x[2],&y[2]);
scanf("%lf%lf",&x[3],&y[3]);
scanf("%lf%lf",&x[4],&y[4]);
int n=0;
p[n++]=Point(x[3],y[3]);
p[n++]=Point(x[4],y[3]);
p[n++]=Point(x[4],y[4]);
p[n++]=Point(x[3],y[4]);
int m=0;
q[m++]=Point(x[1],y[1]);
q[m++]=Point(x[1],y[2]);
q[m++]=Point(x[2],y[1]);
tubao(q,m,ch);
int cnt=0;
for(int i=0;i<n;i++){
if(intubao(ch,m,p[i])){
ans[cnt++]=p[i];
}
}
for(int i=0;i<m;i++){
if(intubao(p,n,q[i])){
ans[cnt++]=q[i];
}
}
p[n]=p[0];
q[m]=q[0];
for(int i=0;i<n;i++){
for(int j=0;j<m;j++){
Point A=line_line(p[i],p[i+1],q[j],q[j+1]);
if(dcmp(A.x-123.456)!=0)ans[cnt++]=A;
}
}
int n1=tubao(ans,cnt,ch);
if(n1<=2)printf("%.8lf\n",0.0);
else printf("%.8lf\n",tubaos(ch,n1));
}
}
字符串双哈希
https://codeforc.es/contest/1200/problem/E
题意：给定一堆字符串，依次插入答案串尾部，每次删掉答案串的后缀 与 待插入串的前缀的最大匹配串
思路：hash匹配后缀和前缀，不断更新hash值
#define bug(x) cout<<#x<<" is "<<x<<endl
#define IO std::ios::sync_with_stdio(0)
#include <bits/stdc++.h>
#define iter ::iterator
#define pa pair<int,int>
using namespace std;
#define ll long long
#define mk make_pair
#define pb push_back
#define se second
#define fi first
#define ls o<<1
#define rs o<<1|1
const int N=1e6+5;
int n;
ll mod[5]={1000000009,998244353};
ll base[5]={666127,777131};
ll p[N][2];
ll h[N][2];
void init(){
for(int i=0;i<2;i++)p[0][i]=1;
for(int i=1;i<=N-2;i++){
for(int j=0;j<2;j++){
p[i][j]=p[i-1][j]*base[j]%mod[j];
}
}
}
ll cal(int l,int r,int t){
return (h[r][t]-h[l-1][t]*p[(r-l+1)][t]%mod[t]+mod[t])%mod[t];
}
ll get(char c){
if('a'<=c&&c<='z')return c-'a';
if('A'<=c&&c<='Z')return c-'A'+26;
if('0'<=c&&c<='9')return c-'0'+52;
}
string s,t;
int main(){
IO;
cin>>n;
cin>>s;
int pre=1;
init();
for(int g=1;g<n;g++){
cin>>t;
int ns=s.length();
for(int i=pre;i<=ns;i++){
ll c=get(s[i-1]);
for(int j=0;j<2;j++){
h[i][j]=(h[i-1][j]*base[j]%mod[j]+c)%mod[j];
}
}
int nt=t.length();
ll H[2]={0};
pre=ns+1;
int id=0;
for(int i=1;i<=nt;i++){
int f=0;
ll c=get(t[i-1]);
for(int j=0;j<2;j++){
H[j]=(H[j]*base[j]%mod[j]+c)%mod[j];
if(cal(ns-i+1,ns,j)==H[j])f++;
}
if(f==2){
id=i;
}
if(i==ns)break;
}
s+=t.substr(id,nt-id);
}
cout<<s<<endl;
}
/*
7
RAtONC14epz KfIenADgDKDci OMmOOQOc yVrfGLV49fW1 xntodZLM5 2f7LXdzX xIhm
5
I want to order pizza
*/
离散化和部分常用STL操作
sort(a,a+n);
int m=unique(a,a+n)-a;
for(int i=0;i<n;i++) b[i]=lower_bound(a,a+m,b[i])-a+1;
lower_bound(v.begin(), v.end(), x)//vector的lower_bound用法
multiset<int>s;
s.insert(x);
s.erase(x)//既可以删除与x相等的所有元素，也可以删除单个迭代器
s.erase(L,R)//删除迭代器[L,R)内的所有元素，是左闭右开区间。
auto id=s.lower_lound(x);
auto id=s.upper_bound(x);
List双向链表
#include <bits/stdc++.h>
using namespace std;
int main() {
list<int> list1 = {1, 2, 3, 4, 5}; // 创建两个 list
list<int> list2 = {10, 20, 30, 40, 50};
list1.splice(list1.begin(), list2); // 1. 使用 splice 移动整个 list2 到 list1 的开始位置
//list1: 10 20 30 40 50 1 2 3 4 5
//list2: 为空
auto it=find(list1.begin(),list2.end(),3);//指定list1某个位置移动到list1最前面
list1.splice(list1.begin(),list1,it);
//3 10 20 30 40 50 1 2 4 5
// 3. 使用 splice 移动 list1 中的一段元素 [20, 40] 到 list2 的结束位置
auto first = find(list1.begin(), list1.end(), 20);
auto last = find(list1.begin(), list1.end(), 40);
if (first != list1.end() && last != list1.end()) {
last = next(last); // 使 last往后移动一次，再将[fisrt,last)区间内的数移到指定位置
list2.splice(list2.end(), list1, first, last);//将list1的[first,last]中元素移动到list2末尾
}
// list1: 3 10 50 1 2 4 5
// list2: 20 30 40
it = find(list1.begin(), list1.end(), 50);
auto it1=next(it); // 得到it的下一个迭代器，指向的数为1
auto it2=prev(it); // 得到it的上一个迭代器，指向的数为10
cout<<*it1<<" "<<*it2<<endl;
}
标签：return,Point,int,ACM,Vector,mod,模板,define From： https://www.cnblogs.com/ccsu-kid/p/18226374