求第k大用堆維護最值並出堆的時候擴展的經典題...

　　因為只有正數，所以一個矩陣的權值肯定比它的任意子矩陣的權值大，那麽一開始把所有滿足條件的最小矩陣加進堆裏，彈出的時候上下左右擴展一行加進堆，用hash判重即可

#include<iostream>
#include<cstring>
#include<cstdlib>
#include<cstdio>
#include<queue>
#define uint unsigned int
#define ull unsigned long long
using
 
 namespace std;
const int maxn=1010, mod=10000019;
struct poi{int x1, yy1, x2, y2; uint sum;};
struct tjm{ull too; int pre;}e[2000010];
priority_queue<poi>q;
bool operator<(poi a, poi b){return a.sum>b.sum;}
int n, m, mna, mnb, k, tot;
int last[mod];
uint mp[maxn][maxn];
void read(uint &k)
{
    
 
int f=1; k=0; char c=getchar();
    while(c<‘0‘ || c>‘9‘) c==‘-‘ && (f=-1), c=getchar();
    while(c<=‘9‘ && c>=‘0‘) k=k*10+c-‘0‘, c=getchar();
    k*=f;
}
inline void add(int x, ull y){e[++tot].too=y; e[tot].pre=last[x]; last[x]=tot;}
inline int makehs(ull x)
{
    int
 
 tmp=x%mod;
    for(int i=last[tmp];i;i=e[i].pre)
    if(e[i].too==x) return i;
    add(tmp, x);
    return tot;
}
inline int hs(ull x)
{
    int tmp=x%mod;
    for(int i=last[tmp];i;i=e[i].pre)
    if(e[i].too==x) return i;
    return -1;
}
inline void update(int x1, int yy1, int x2, int y2)
{
    ull tmp=0;
    tmp=tmp*2333+x1; tmp=tmp*2333+yy1; tmp=tmp*2333+x2; tmp=tmp*2333+y2;
    if(hs(tmp)!=-1) return;
    makehs(tmp);
    q.push((poi){x1, yy1, x2, y2, mp[x2][y2]-mp[x2][yy1-1]-mp[x1-1][y2]+mp[x1-1][yy1-1]});
}
int main()
{
    scanf("%d%d%d%d%d", &n, &m, &mna, &mnb, &k);
    for(int i=1;i<=n;i++)
        for(int j=1;j<=m;j++)
            read(mp[i][j]), mp[i][j]+=mp[i-1][j]+mp[i][j-1]-mp[i-1][j-1];
    for(int i=mna;i<=n;i++)
        for(int j=mnb;j<=m;j++)
        {
            ull tmp=0;
            int x1=i-mna+1, yy1=j-mnb+1, x2=i, y2=j;
            tmp=tmp*2333+x1; tmp=tmp*2333+yy1; tmp=tmp*2333+x2; tmp=tmp*2333+y2;
            makehs(tmp);
            q.push((poi){x1, yy1, x2, y2, mp[x2][y2]-mp[x2][yy1-1]-mp[x1-1][y2]+mp[x1-1][yy1-1]});
        }
    for(int i=1;i<=k;i++)
    {
        if(q.empty()) return puts("-1"), 0;
        poi t=q.top(); q.pop();
        int x1=t.x1, yy1=t.yy1, x2=t.x2, y2=t.y2;
        if(x1>1) update(x1-1, yy1, x2, y2);
        if(yy1>1) update(x1, yy1-1, x2, y2);
        if(x2<n) update(x1, yy1, x2+1, y2);
        if(y2<m) update(x1, yy1, x2, y2+1);
        if(i==k) printf("%u\n", t.sum);
    }
}

bzoj4165: 矩陣（堆+hash）