最大正方形

Given a 2D binary matrix filled with 0's and 1's, find the largest square containing all 1's and return its area.

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Example

For example, given the following matrix:

1 0 1 0 0
1 0 1 1 1
1 1 1 1 1
1 0 0 1 0

Return 4.

判断：dp[i][j]为最大正方形的右下角，则dp[i-1][j-1],dp[i-1][j],dp[i][j-1]是几个正方形的右下角。否则，dp[i][j]所属的正方形是自己。

正方形的边长：dp[i][j]是dp[i-1][j],dp[i-1][j-1],dp[i][j-1]中最小+1.

即dp[i][j]=min(dp[i-1][j-1],dp[i-1][j],dp[i][j-1])+1;

#include <iostream>
using namespace std;
int dp[1005][1005],g[1005][1005];
int hs(int x,int y)
{
	int s,i,j;
	s=0;
	for(i=0;i<x;i++)
		for(j=0;j<y;j++)
		{
			dp[i][j]=(g[i][j]+1)%2;
			s|=dp[i][j];
		}
	for(i=1;i<x;i++)
		for(j=1;j<y;j++)
		{
			if(g[i][j])
				dp[i][j]=0;
			else
			{
				dp[i][j]=min(dp[i-1][j-1],min(dp[i-1][j],dp[i][j-1]))+1;
				s=max(s,dp[i][j]);
			}
		}
	return s*s;
}
int main()
{
	int h,w;
	int i,j;
	cin>>h>>w;
	for(i=0;i<h;i++)
		for(j=0;j<w;j++)
			cin>>g[i][j];
	cout<<hs(h,w);
	return 0;
}