POJ 3494 Largest Submatrix of All 1’s（单调栈）

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.
Input
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 lines each with n numbers. The input ends once EOF is met.
Output
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.
Sample Input
2 2
0 0
0 0
4 4
0 0 0 0
0 1 1 0
0 1 1 0
0 0 0 0
Sample Output
0
4

首先找每个点向上的最大高度，然后再对每个点向左扩展，再向右扩展。
找每个点向左向右方向第一个比它小的位置是一个单调减栈。

#include<iostream>
#include<cstdio>
#include<stack>
#include <cstring>
#include<algorithm>
using namespace std;

int m,n;
int a[2005][2005];
int u[2005][2005];
int l[2005][2005],r[2005][2005];

int main()
{
   
    while(scanf("%d%d",&m,&n)!=EOF){
   
        for(int i=1;i<=m;i++){
   
            for(int j=1;j<=n;j++){
   
                scanf("%d",&a[i][j]);
            }
        }
        stack<pair<int,int> > s;
        memset(u,0,sizeof(u));
        for(int i=1;i<=m;i++){
   //求出每个点向上连续的最大长度 
            for(int j=1;j<=n;j++){
   
                if(a[i][j]==0) u[i][j]=0;
                else u[i][j]=u[i-1][j]+1;
            }
        }
        for(int i=1;i<=m;i++){
   
            while(!s.empty()) s.pop();
            for(int j=1;j<=n;j++){
   
                while(!s.empty()&&s.top().first>=u[i][j]){
   //找左边第一个比它小的位置 
                    s.pop();
                }
                if(s.empty()) l[i][j]=1;
                else l[i][j]=s.top().second+1;
                s.push(pair<int,int>(u[i][j],j));
            }
        }

            for(int i=1;i<=m;i++){
   
                while(!s.empty()) s.pop();
                for(int j=n;j>=1;j--){
   
                    while(!s.empty()&&s.top().first>=u[i][j]){
   
                    s.pop();
                    }
                    if(s.empty()) r[i][j]=m;
                    else r[i][j]=s.top().second-1;
                    s.push(pair<int,int>(u[i][j],j));
             }
            }

            int res=0;
            for(int i=1;i<=m;i++){
   
                for(int j=1;j<=n;j++){
   
                    res=max(res,u[i][j]*(r[i][j]-l[i][j]+1));
                }
            }
        printf("%d\n",res);
    }
    return 0;
}