UVALive 4867 Maximum Square 贪心

E - Maximum Square

Time Limit:4500MS     Memory Limit:0KB     64bit IO Format:%lld & %llu

Submit Status Practice UVALive 4867

Description

Given an NxM matrix of all 1s and 0s, find the largest submatrix which is a square containing all 1s.

Input

There will be several test cases in the input. Each test case will begin with two integers, N and M(1N, M1, 000) indicating the number of rows and columns of the matrix. The next N lines will each contain M space-separated integers, guaranteed to be either 0 or 1. The input will end with a line with two 0s.

Output

For each test case, print a single integer, indicating the width (and height) of the largest square of all 1s, or 0 if there are no 1s. Print no extra spaces, and do not print any blank lines between answers.

Sample Input

4 5 
0 1 0 1 1 
1 1 1 1 1 
0 1 1 1 0 
1 1 1 1 1 
3 4 
1 1 1 1 
1 1 1 1 
1 1 1 1 
6 6 
0 0 0 0 0 0 
0 0 0 0 0 0 
0 0 0 0 0 0 
0 0 0 0 0 0 
0 0 0 0 0 0 
0 0 0 0 0 0 
0 0

Sample Output

3 
3 
0

#include <cstdio>
#include <cmath>
#include <cstring>
#include <ctime>
#include <iostream>
#include <algorithm>
#include <set>
#include <vector>
#include <sstream>
#include <queue>
#include <typeinfo>
#include <fstream>
typedef long long ll;
using namespace std;
//freopen("D.in","r",stdin);
//freopen("D.out","w",stdout);
#define sspeed ios_base::sync_with_stdio(0);cin.tie(0)
#define maxn 1010
const int inf=0x7fffffff;   //无限大
int g[maxn][maxn];
int main()
{
    int m,n;
    while(cin>>n>>m)
    {
        if(m==0&&n==0)
            break;
        memset(g,0,sizeof(g));
        for(int i=0;i<n;i++)
        {
            for(int j=0;j<m;j++)
            {
                cin>>g[i][j];
            }
        }
        int ans=0;
        for(int i=0;i<n;i++)
        {
            for(int j=0;j<m;j++)
            {
                if(i==0||j==0)
                {
                    ans=max(g[i][j],ans);
                    continue;
                }
                if(g[i][j]==0)
                    continue;
                g[i][j]=min(g[i-1][j],min(g[i][j-1],g[i-1][j-1]))+1;
                ans=max(g[i][j],ans);
            }
        }

        cout<<ans<<endl;
    }
    return 0;
}