传送门
Solution
考虑看到什么k小就整体二分套上去试一下。
矩形k小整体二分+二维树状数组就好了。
代码实现
// luogu-judger-enable-o2
/*
mail: mleautomaton@foxmail.com
author: MLEAutoMaton
This Code is made by MLEAutoMaton
*/
#include<stdio.h>
#include<stdlib.h>
#include<string.h>
#include<math.h>
#include<algorithm>
#include<queue>
#include<set>
#include<map>
#include<iostream>
using namespace std;
#define ll long long
#define re register
#define file(a) freopen(a".in","r",stdin);freopen(a".out","w",stdout)
inline int gi()
{
int f=1,sum=0;char ch=getchar();
while(ch>'9' || ch<'0'){if(ch=='-')f=-1;ch=getchar();}
while(ch>='0' && ch<='9'){sum=(sum<<3)+(sum<<1)+ch-'0';ch=getchar();}
return f*sum;
}
const int N=1000010;
struct node
{
int x,y,k,xx,yy,id;
}q[N],q1[N],q2[N];
int c[510][510],n,Q,ans[N];
int lowbit(int x){return x&(-x);}
void Add(int x,int y,int d){for(int i=x;i<=n;i+=lowbit(i))for(int j=y;j<=n;j+=lowbit(j))c[i][j]+=d;}
int query(int x,int y){int ret=0;for(int i=x;i>0;i-=lowbit(i))for(int j=y;j>0;j-=lowbit(j))ret+=c[i][j];return ret;}
int query(int x,int xx,int y,int yy){return query(xx,yy)-query(x-1,yy)-query(xx,y-1)+query(x-1,y-1);}
void solve(int l,int r,int ql,int qr)
{
if(ql>qr)return;
if(l==r)
{
for(re int i=ql;i<=qr;i++)
if(q[i].id)ans[q[i].id]=l;
return;
}
int mid=(l+r)>>1,L=0,R=0;
for(re int i=ql;i<=qr;i++)
if(q[i].id)
{
int sum=query(q[i].x,q[i].xx,q[i].y,q[i].yy);
if(sum>=q[i].k)q1[++L]=q[i];
else{q[i].k-=sum;q2[++R]=q[i];}
}
else
{
if(q[i].k<=mid){q1[++L]=q[i];Add(q[i].x,q[i].y,1);}
else q2[++R]=q[i];
}
for(re int i=ql;i<=qr;i++)
if(!q[i].id && q[i].k<=mid)Add(q[i].x,q[i].y,-1);
for(re int i=1;i<=L;i++)q[ql+i-1]=q1[i];
for(re int i=1;i<=R;i++)q[ql+L+i-1]=q2[i];
solve(l,mid,ql,ql+L-1);solve(mid+1,r,ql+L,qr);
}
int main()
{
n=gi();int cnt=gi();
for(re int i=1;i<=n;i++)
for(re int j=1;j<=n;j++)
q[++Q]=(node){i,j,gi(),0,0,0};
int Time=0;
while(cnt--)
{
Time++;
int x=gi(),y=gi(),xx=gi(),yy=gi(),k=gi();
q[++Q]=(node){x,y,k,xx,yy,Time};
}
solve(1,1e9,1,Q);
for(int i=1;i<=Time;i++)
printf("%d
",ans[i]);
return 0;
}