POJ - 3494 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

（题意：求出一个最大全为1的子矩阵的面积。）

（思路：枚举每一行，用单调栈正序和倒序递增维护每一列的“高度”，用l和r数组记下能到达的列，取最大值即可。）

#include 
#include 
using namespace std;
const int N = 2e3+10;
int h[N],l[N],r[N],sta[N],top;
int n,m;
int main(void)
{
	while(~scanf("%d%d",&n,&m))	
	{
		int x,ans=0;
		for(int i=1;i<=m;i++) h[i]=0;
		for(int i=1;i<=n;i++)
		{
			for(int j=1;j<=m;j++)
			{
				scanf("%d",&x);
				h[j]=x?h[j]+1:0;
			}
			//cout<<"-------------------"<0&&h[sta[top]]>=h[k])	
					top--;
				if(top==0) l[k]=1;
				else l[k]=sta[top]+1;
				sta[++top]=k;
			}
			top=0;
			for(int k=m;k>0;k--)
			{
				while(top>0&&h[sta[top]]>=h[k])	
					top--;
				if(top==0) r[k]=n;
				else r[k]=sta[top]-1;
				sta[++top]=k;
			}
			
			for(int k=1;k<=m;k++)
			{
				ans=max(ans,h[k]*(r[k]-l[k]+1));
			//	cout<