POJ - 3494 Largest Submatrix of All 1’s 單調棧求最大子矩陣

Largest Submatrix of All 1’s

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 mlines 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


很巧妙的方法。把圖拆成一行行來作，分別求以每行爲底的最大矩形面積，相似lc課後輔導。最後比較出最大值便可。

#include<stdio.h>
#include<stack>
using namespace std;

int a[2005][2005];
int l[2005],r[2005];

int main()
{
    int n,m,max,i,j;
    while(~scanf("%d%d",&n,&m)){
        max=0;
        for(i=1;i<=n;i++){
            for(j=1;j<=m;j++){
                scanf("%d",&a[i][j]);
                if(a[i][j]==1) a[i][j]+=a[i-1][j];
            }
        }
        for(i=1;i<=n;i++){
            stack<int> s;
            for(j=1;j<=m;j++){
                while(s.size()&&a[i][s.top()]>=a[i][j]) s.pop();
                l[j]=s.size()==0?1:s.top()+1;
                s.push(j);
            }
            while(s.size()){
                s.pop();
            }
            for(j=m;j>=1;j--){
                while(s.size()&&a[i][s.top()]>=a[i][j]) s.pop();
                r[j]=s.size()==0?m:s.top()-1;
                s.push(j);
            }
            for(j=1;j<=m;j++){
                if(a[i][j]*(r[j]-l[j]+1)>max) max=a[i][j]*(r[j]-l[j]+1);
            }
        }
        printf("%d\n",max);
    }
    return 0;
}