P3870 [TJOI2009] 开关
先看一道经典的区间开关灯问题的模型,维护一个lz 每次异或操作就好了
#include<bits/stdc++.h>
using namespace std;
using ll = long long;
using pii = pair<int,int>;
const int N = 1e5+10;
const int inf = 0x3f3f3f3f;
const int mod = 1e9+7;
int gcd(int a,int b){return b?gcd(b,a%b):a;}
int lcm(int a,int b){return a*b/gcd(a,b);}
int qmi(int a,int b,int mod){int res=1;while(b){if(b&1)res=res*a%mod;b>>=1;a=a*a%mod;}return res;}
int n,q,m;
// int e[N],ne[N],w[N],h[N],idx;
// void add(int a,int b,int c){
// e[idx] = b,ne[idx] = h[a],w[idx] = c,h[a] = idx++;
// }
struct Segment{
int l,r;
int lz;
int s;
}tr[N<<2];
void pushup(int u){
tr[u].s = tr[u<<1].s + tr[u<<1|1].s;
}
void pushdown(int u){
if(tr[u].lz){
tr[u<<1].lz^=1;
tr[u<<1|1].lz^=1;
tr[u<<1].s = tr[u<<1].r-tr[u<<1].l+1-tr[u<<1].s;
tr[u<<1|1].s = tr[u<<1|1].r-tr[u<<1|1].l+1-tr[u<<1|1].s;
tr[u].lz = 0;
}
}
void build(int u,int l,int r){
tr[u] = {l,r,0,0};
if(l==r)return;
int mid = (l+r)/2;
build(u<<1,l,mid),build(u<<1|1,mid+1,r);
pushup(u);
}
void modify(int u,int l,int r){
if(l<=tr[u].l&&tr[u].r<=r){
tr[u].lz^=1;
tr[u].s = tr[u].r-tr[u].l+1-tr[u].s;
return;
}
pushdown(u);
int mid = (tr[u].l+tr[u].r)/2;
if(l<=mid)modify(u<<1,l,r);
if(r>mid)modify(u<<1|1,l,r);
pushup(u);
}
int query(int u,int l,int r){
if(l<=tr[u].l&&tr[u].r<=r){
return tr[u].s;
}
pushdown(u);
int res = 0;
int mid = (tr[u].l+tr[u].r)/2;
if(l<=mid)res+=query(u<<1,l,r);
if(r>mid)res+=query(u<<1|1,l,r);
return res;
}
void solve()
{
cin>>n>>q;
build(1,1,n);
while(q--){
int op,l,r;cin>>op>>l>>r;
if(!op)modify(1,l,r);
else cout<<query(1,l,r)<<"\n";
}
}
signed main()
{
ios::sync_with_stdio(0),cin.tie(0),cout.tie(0);
int _;
//cin>>_;
_ = 1;
while(_--)solve();
return 0;
}
E. Danil and a Part-time Job
再看一道这个问题的树上版本
结合一下dfn序处理区间问题就好了
#include<bits/stdc++.h>
using namespace std;
using ll = long long;
using pii = pair<int,int>;
const int N = 2e5+10;
const int inf = 0x3f3f3f3f;
const int mod = 1e9+7;
int gcd(int a,int b){return b?gcd(b,a%b):a;}
int lcm(int a,int b){return a*b/gcd(a,b);}
int qmi(int a,int b,int mod){int res=1;while(b){if(b&1)res=res*a%mod;b>>=1;a=a*a%mod;}return res;}
int n,q,m;
// int e[N],ne[N],w[N],h[N],idx;
// void add(int a,int b,int c){
// e[idx] = b,ne[idx] = h[a],w[idx] = c,h[a] = idx++;
// }
int cost[N];
int dfn[N],in[N],out[N],times;
vector<int>g[N];
struct Segment{
int l,r;
int lz;
int s;
}tr[N<<2];
void dfs(int u){
in[u] = ++times;
dfn[times] = u;
for(int &t:g[u])dfs(t);
out[u] = times;
}
void pushup(int u){
tr[u].s = tr[u<<1].s + tr[u<<1|1].s;
}
void pushdown(int u){
if(tr[u].lz){
tr[u<<1].lz^=1;
tr[u<<1|1].lz^=1;
tr[u<<1].s = tr[u<<1].r-tr[u<<1].l+1-tr[u<<1].s;
tr[u<<1|1].s = tr[u<<1|1].r-tr[u<<1|1].l+1-tr[u<<1|1].s;
tr[u].lz = 0;
}
}
void build(int u,int l,int r){
tr[u] = {l,r};
if(l==r){
tr[u].lz = 0;
tr[u].s = cost[dfn[l]];
return;
}
int mid = (l+r)/2;
build(u<<1,l,mid),build(u<<1|1,mid+1,r);
pushup(u);
}
void modify(int u,int l,int r){
if(l<=tr[u].l&&tr[u].r<=r){
tr[u].lz^=1;
tr[u].s = tr[u].r-tr[u].l+1-tr[u].s;
return;
}
pushdown(u);
int mid = (tr[u].l+tr[u].r)/2;
if(l<=mid)modify(u<<1,l,r);
if(r>mid)modify(u<<1|1,l,r);
pushup(u);
}
int query(int u,int l,int r){
if(l<=tr[u].l&&tr[u].r<=r){
return tr[u].s;
}
pushdown(u);
int res = 0;
int mid = (tr[u].l+tr[u].r)/2;
if(l<=mid)res+=query(u<<1,l,r);
if(r>mid)res+=query(u<<1|1,l,r);
return res;
}
void solve()
{
cin>>n;
for(int i=2;i<=n;++i){int fa;cin>>fa;g[fa].push_back(i);}
dfs(1);
for(int i=1;i<=n;i++)cin>>cost[i];
build(1,1,n);
cin>>q;
while(q--){
string op;cin>>op;
int id;cin>>id;
if(op=="get")cout<<query(1,in[id],out[id])<<"\n";
else modify(1,in[id],out[id]);
}
}
signed main()
{
ios::sync_with_stdio(0),cin.tie(0),cout.tie(0);
int _;
//cin>>_;
_ = 1;
while(_--)solve();
return 0;
}
标签:return,idx,int,res,模型,开关,区间,const,mod
From: https://blog.csdn.net/m0_60921016/article/details/137115706