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;

}