题目链接 BZOJ
洛谷

整体二分。把求序列第K小的树状数组改成二维树状数组就行了。
初始答案区间有点大，离散化一下。

因为这题是一开始给点，之后询问，so可以先处理该区间值在l~mid的修改，再处理询问。即二分标准可以直接用点的标号。
结构体的赋值可以改为赋值操作的编号。（这样内存没那么连续？想多了你）

改了半下午，优化了500ms。。

//6980kb    10584ms 好慢啊QAQ
//4208ms    7.59MB
#include <cstdio>
#include <cctype>
#include <algorithm>
#define lb(x) ((x)&-(x))
//#define gc() getchar()
#define MAXIN 500000
#define gc() (SS==TT&&(TT=(SS=IN)+fread(IN,1,MAXIN,stdin),SS==TT)?EOF:*SS++)
const int N=505,M=60005;int n,m,Ans[M],q[M],q1[M],q2[M];
char IN[MAXIN],*SS=IN,*TT=IN;
struct Point
{int x,y,val;Point() {}Point(int x,int y,int val):x(x),y(y),val(val) {}bool operator <(const Point &a)const{return val<a.val;}
}pt[N*N];
inline int read();
struct Operation//Query
{int K,x1,y1,x2,y2;inline void Input(){x1=read(),y1=read(),x2=read(),y2=read(),K=read();}
}op[M];inline int read()
{int now=0;register char c=gc();for(;!isdigit(c);c=gc());for(;isdigit(c);now=now*10+c-'0',c=gc());return now;
}
namespace T
{int n,t[N][N];inline void Modify(int x,int y,int v){for(int i=x; i<=n; i+=lb(i))for(int j=y; j<=n; j+=lb(j)) t[i][j]+=v;}inline void Clear(int x,int y){for(int i=x; i<=n; i+=lb(i))for(int j=y; j<=n; j+=lb(j))if(t[i][j]) t[i][j]=0; else break;}inline int Query(int x,int y){int res=0;for(int i=x; i; i^=lb(i))for(int j=y; j; j^=lb(j)) res+=t[i][j];return res;}inline int Query_Area(Operation q){//prefix sumreturn Query(q.x2,q.y2)-Query(q.x1-1,q.y2)-Query(q.x2,q.y1-1)+Query(q.x1-1,q.y1-1);}
}
void Solve(int l,int r,int h,int t)
{if(h>t) return;if(l==r){for(int i=h; i<=t; ++i) Ans[q[i]]/*[op[q[i]].pos]*/=pt[l].val;return;}int mid=l+r>>1, t1=0, t2=0;for(int i=l; i<=mid; ++i) T::Modify(pt[i].x,pt[i].y,1);for(int now,tmp,i=h; i<=t; ++i){now=q[i], tmp=T::Query_Area(op[now]);if(tmp>=op[now].K) q1[t1++]=now;else op[now].K-=tmp, q2[t2++]=now;}for(int i=l; i<=mid; ++i) T::Clear(pt[i].x,pt[i].y);for(int i=0; i<t1; ++i) q[h+i]=q1[i];for(int i=0; i<t2; ++i) q[h+t1+i]=q2[i];Solve(l,mid,h,h+t1-1), Solve(mid+1,r,h+t1,t);
}int main()
{T::n=n=read(), m=read();int tot=0;for(int i=1; i<=n; ++i)for(int j=1; j<=n; ++j) pt[++tot]=Point(i,j,read());std::sort(pt+1,pt+1+tot);for(int i=1; i<=m; ++i) q[i]=i, op[i].Input();Solve(1,tot,1,m);for(int i=1; i<=m; ++i) printf("%d\n",Ans[i]);return 0;
}

优化前：

//4680ms    17.75MB
#include <cstdio>
#include <cctype>
#include <algorithm>
#define lb(x) ((x)&-(x))
//#define gc() getchar()
#define MAXIN 60000
#define gc() (SS==TT&&(TT=(SS=IN)+fread(IN,1,MAXIN,stdin),SS==TT)?EOF:*SS++)
const int N=505,M=60005+N*N;int n,m,Q,cnt,A[N*N],Ans[60005];
char IN[MAXIN],*SS=IN,*TT=IN;
struct Operation
{int K,x1,y1,x2,y2,pos;//K=0: Modify (x1,y1):=posOperation() {}Operation(int K,int x1,int y1,int x2,int y2,int pos):K(K),x1(x1),y1(y1),x2(x2),y2(y2),pos(pos) {}
}q[M],q1[M],q2[M];inline int read()
{int now=0;register char c=gc();for(;!isdigit(c);c=gc());for(;isdigit(c);now=now*10+c-'0',c=gc());return now;
}
namespace T
{int n,t[N][N];inline void Modify(int x,int y,int v){for(int i=x; i<=n; i+=lb(i))for(int j=y; j<=n; j+=lb(j)) t[i][j]+=v;}inline void Clear(int x,int y){for(int i=x; i<=n; i+=lb(i))for(int j=y; j<=n; j+=lb(j))if(t[i][j]) t[i][j]=0; else break;}inline int Query(int x,int y){int res=0;for(int i=x; i; i^=lb(i))for(int j=y; j; j^=lb(j)) res+=t[i][j];return res;}inline int Query_Area(Operation q){//prefix sumreturn Query(q.x2,q.y2)-Query(q.x1-1,q.y2)-Query(q.x2,q.y1-1)+Query(q.x1-1,q.y1-1);}
}
void Solve(int l,int r,int h,int t)
{if(h>t) return;if(l==r){for(int i=h; i<=t; ++i) if(q[i].K) Ans[q[i].pos]=A[l];return;}bool goon=0;for(int i=h; i<=t; ++i) if(q[i].K) {goon=1; break;}if(!goon) return;int mid=l+r>>1, midV=A[mid], t1=0, t2=0;for(int i=h; i<=t; ++i)if(q[i].K){int tmp=T::Query_Area(q[i]);//这样好像少做几次加法！但是多copy两个int。。（你够了→_→）if(tmp>=q[i].K) q1[t1++]=q[i];else q[i].K-=tmp, q2[t2++]=q[i];}else{if(q[i].pos<=midV) T::Modify(q[i].x1,q[i].y1,1), q1[t1++]=q[i];else q2[t2++]=q[i];}for(int i=0; i<t1; ++i) if(!q1[i].K) T::Clear(q1[i].x1,q1[i].y1);for(int i=0; i<t1; ++i) q[h+i]=q1[i];for(int i=0; i<t2; ++i) q[h+t1+i]=q2[i];Solve(l,mid,h,h+t1-1), Solve(mid+1,r,h+t1,t);
}int main()
{T::n=n=read(), m=read(), Q=0;for(int i=1; i<=n; ++i)for(int j=1; j<=n; ++j) q[++Q]=Operation(0,i,j,0,0,A[Q]=read());std::sort(A+1,A+1+Q), cnt=1;for(int i=2; i<=Q; ++i) if(A[i]!=A[i-1]) A[++cnt]=A[i];for(int x1,y1,x2,y2,i=1; i<=m; ++i)x1=read(),y1=read(),x2=read(),y2=read(),q[++Q]=Operation(read(),x1,y1,x2,y2,i);Solve(1,cnt,1,Q);for(int i=1; i<=m; ++i) printf("%d\n",Ans[i]);return 0;
}