Given a m-by-n (0,1)-matrix, of all its submatrices of all 1’s which is the largest? By largest we mean that the submatrix has the most elements.

The input contains multiple test cases. Each test case begins with m and n (1 ≤ m, n ≤ 2000) on line. Then come the elements of a (0,1)-matrix in row-major order on m

For each test case, output one line containing the number of elements of the largest submatrix of all 1’s. If the given matrix is of all 0’s, output 0.

2 2
0 0
0 0
4 4
0 0 0 0
0 1 1 0
0 1 1 0
0 0 0 0

0
4
 


單調棧，每一行記錄個列的高度 然後每一行進行單調棧，用輸入輸出流還是超時得用scanf。。或者 用scanf就不錯。。

#include <iostream>
#include <cstdio>
#include <cstring>
#include <map>
#include <queue>
#include <cmath>
using namespace std;
int n,m,d;
int mp[2002][2002];
int l[2002]={1},r[2002],stack[2001],top,maxi,sum;
 
int main()
{
    //ios::sync_with_stdio(false);
    //cin.tie(0);

    while(scanf("%d%d",&n,&m)!=EOF)
    {
        maxi=0;
        for(int i=1;i<=n;i++)
        {
            top=1;
            for(int j=1;j<=m;j++)
            {
                scanf("%d",&d);
                //mp[i][j]=mp[i][j-1]+d;  不對的 中間如果有0 應該中斷
                mp[i][j]=(d==0?0:(mp[i-1][j]+d));
                l[j]=r[j]=j;
            }
            for(int j=1;j<=m+1;j++)
            {
                while(top>1&&mp[i][j]<=mp[i][stack[top-1]])
                {
                    r[stack[top-1]]=j-1;
                    sum=(r[stack[top-1]]-l[stack[top-1]]+1)*mp[i][stack[top-1]];
                    if(sum>maxi)maxi=sum;
                    top--;
                }
                l[j]=stack[top-1]+1;
                stack[top++]=j;
            }
        }

        printf("%d\n",maxi);
    }
}
/*
4 4
1 0 1 1
0 1 1 1
0 0 1 1
0 1 1 1
*/

Largest Submatrix of All 1’s