传送门

洛谷

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.comauthor: MLEAutoMatonThis 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\n",ans[i]);return 0;
}