A. A Hero Named Magnus
签到题 输出2x-1 注意用到unsigned long long
I. PTSD
签到题
G. Occupy the Cities
分析：
考虑从左到右依次考虑 pre表示当前1需要向前贡献多少 并且不断跟新pre
因为在有限的时间内 所有的1都尽可能的最大化 想到用二分时间mid
如果遇到pre>mid 这样一定不成立
如果遇到pre=mid 当前1一定要先先前扩展 再向后扩展 贡献为mid-1
如果遇到pre<mid 这样我可以第一次就向后扩展 贡献为mid
细节挺多的 调了好久
#include<bits/stdc++.h>
using namespace std;
#define lowbit(x) x&(-x)
#define ll long long
int ii,pre,edd,sum,T,n,mid,l,r,ans;
string s;
void solve();
bool ck(int x){
ii=0;pre=0;edd=0;
while(ii<n){
sum=0;
while(s[ii]=='0'&&ii<n)ii++,sum++;
if(ii==n){
if(sum>edd)return false;
return true;
}
pre=sum-edd;
if(pre<0)pre=0;
if(pre>x)return false;
if(x>pre)
edd=x;
else edd=x-1;
if(edd<0)edd=0;
ii++;
pre=0;
}
if(pre>0)return false;
else return true;
}
int main(){
cin>>T;
while(T--)solve();
return 0;
}
void solve(){
cin>>n>>s;
l=0,r=n;
while(l<=r){
mid=l+r>>1;
if(ck(mid))r=mid-1,ans=mid;
else l=mid+1;
}
printf("%d\n",ans);
}
E. Buy and Delete
分析： Bob最多删两次边 问题转化为求有向图的最小环
太菜了 有向图找环的问题困扰了我们好久
有向图找最小环
对每个点 i 进行dj最短路 先将i扩展的点放入队列中 再将dis[i]设为inf 最终dis[i]即为i点环的最小值
第一次遇到这样的做法 学到了
#include <iostream>
#include <cstdio>
#include <cstring>
#include <cmath>
#include <algorithm>
#include <vector>
#include <queue>
#include <map>
using namespace std;
typedef long long lld;
const int N = 2005;
const int M = 5005;
const lld inf = 1e17;
int n, m;
lld dis[N], ans, minc = inf, c, w[M];
bool vis[N];
int nextt[M], to[M], head[N], cnt;
void add(int x, int y, lld z) {
nextt[++cnt] = head[x]; w[cnt] = z;
to[cnt] = y; head[x] = cnt;
}
priority_queue<pair<lld, int> > q;
void dijkstra() {
while(!q.empty()) {
int x = q.top().second; q.pop();
if(vis[x]) continue;
vis[x] = true;
for(int i = head[x]; i; i = nextt[i]) {
int y = to[i];
if(dis[y] > dis[x] + w[i]) {
dis[y] = dis[x] + w[i];
q.push(make_pair(-dis[y], y));
}
}
}
}
int main() {
scanf("%d%d%d", &n, &m, &c);
for(int i = 1, x, y; i <= m; i++) {
lld z;
scanf("%d%d%lld", &x, &y, &z);
add(x, y, z); minc = min(minc, z);
}
ans = inf;
for(int x = 1; x <= n; x++) {
for(int i = 1; i <= n; i++) dis[i] = inf;
while(!q.empty()) q.pop();
memset(vis, false, sizeof(vis));
dis[x] = 0;
for(int i = head[x]; i; i = nextt[i]) {
int y = to[i];
if(dis[y] > dis[x] + w[i]) {
dis[y] = dis[x] + w[i];
q.push(make_pair(-dis[y], y));
}
}
dis[x] = inf;
dijkstra();
ans = min(ans, dis[x]);
}
if(ans == inf) {
if(c >= minc) printf("1");
else printf("0");
} else {
if(c >= ans) printf("2");
else if(c >= minc) printf("1");
else printf("0");
}
return 0;
}
D. Assumption is All You Need
题意：给出两个序列a和b，每次能够在a上交换一个逆序对，问能否构造一组序列将a变成b。
分析：这个题目不简单
发现考虑降序数组不太好 所以想到按照顺序贪心
考虑最大数n 如果 ida[n]>idb[n] 即需要将n换到前面去 发现这样是不可能成立的
对于ida[n]=idb[n] 就不需要再动他了
对于ida[n]<idb[n] 并不是一次性和目标位置交换 因为要尽可能大的数放前面 所以依次和大数交换
#include <bits/stdc++.h>
#include <iostream>
#include <algorithm>
#include <cmath>
#include <cstdio>
#include <vector>
#include <cstring>
#include <map>
#include <stack>
#include <set>
#include <queue>
using namespace std;
typedef long long int ll;
#define rep(i,a,n) for(int i=a;i<n;i++)
#define repn(i,a,n) for(int i=a;i<=n;i++)
#define per(i,a,n) for(int i=a;i>=n;i--)
#define mem(a,b) memset(a,b,sizeof(a))
#define pii pair<int,int>
#define pll pair<ll,ll>
#define pdd pair<long double,long double>
#define vl vector<ll>
#define vi vector<int>
#define x first
#define y second
#define eachit(x,y) for(auto x=y.begin();x!=y.end();x++)
#define each(x,y) for(auto& x:y)
#define fastio ios::sync_with_stdio(0);cin.tie(0);cout.tie(0);
#define all(x) (x).begin(), (x).end()
typedef unsigned long long int ull;
#define pb push_back
#define endl '\n'
ll gcd(ll a,ll b){if(b==0)return abs(a);else return gcd(b,a%b);}
ll lcm(ll a,ll b){return a/gcd(a,b)*b;}
const int mod = 998244353;
const ll INFLL=0x3f3f3f3f3f3f3f3fll;
const int INF=0x3f3f3f3f;
const double pi=acos(-1.0);
ll ksm(ll a,ll b)
{ll rec=1;while(b){if(b%2){rec*=a;rec%=mod;}ll tmp=a%mod;a=tmp*tmp;a%=mod;b>>=1;rec%=mod;}rec%=mod;return rec;}
ull fpow(ull a,ull b)
{ull rec=1;while(b){if(b%2){rec*=a;}ull tmp=a;a=tmp*tmp;b>>=1;}return rec;}
double dpow(double a,ll b)
{double rec=1;while(b){if(b%2){rec*=a;}double tmp=a;a=tmp*tmp;b>>=1;}return rec;}
int lowbit(int lowbitn){return lowbitn&(-lowbitn);}
ll lowbit(ll lowbitn){return lowbitn&(-lowbitn);}
const int N = 2050;
int a[N], b[N];
int ida[N], idb[N];
vector<pii> ans;
signed main()
{
fastio
int t;
cin >> t;
while(t --)
{
ans.clear();
int n;
cin >> n;
repn(i, 1, n)
{
cin >> a[i];
ida[a[i]] = i;
}
repn(i, 1, n)
{
cin >> b[i];
idb[b[i]] = i;
}
bool flag = 1;
per(i, n, 1)
{
if(ida[i] == idb[i]) continue;
if(ida[i] > idb[i])
{
//cout << i << endl;
flag = 0;
break;
}
int now = ida[i], aim = idb[i];
per(j, i - 1, 1)
{
if(ida[j] > now && ida[j] <= aim)
{
ida[i] = ida[j];
swap(a[now], a[ida[j]]);
swap(now, ida[j]);
ans.pb({now, ida[j]});
}
}
}
repn(i, 1, n)
{
if(a[i] != b[i]) flag = 0;
}
if(!flag) cout << -1 << endl;
else
{
cout << ans.size() << endl;
each(it, ans)
{
if(it.x > it.y) swap(it.x, it.y);
cout << it.x << " " << it.y << endl;
}
}
}
return 0;
}