BOJ 1573 Largest Square

Largest Square

Accept:24     Submit:50

Time Limit:5000MS     Memory Limit:65536KB

           

Description

Give you a N*N(1≤N≤2000)Square.For each grid in the square,its color is red or black.Your task is to find a largest red square.

InputFormat

The first line is an integer T(1≤T≤5) indicating the case number.

For each case,the first line is N,then next N lines describes the square.If a_ij=0,indicating a_ijis black,if a_ij=1,indicating a_ij is red.

OutputFormat

For each case,output an integer which is the leng of the side of the largest red square.

SampleInput

2

3

101

000

111

3

111

111

111

SampleOutput

1

3

新生排位赛第三场的E题，比赛时参考了POJ1050的思想，把每一个二维的正方形暴力转化为一维的数组来处理，结果以6003ms的TLE宣告失败。

听学长讲了以后才知道，这题可以动态规划处理，用一个数组dp[i][j]记录以(i,j)为右下角可以构成的最大红色正方形的边长。

具体的状态转移可以参看代码中的f()函数。

#include<iostream>
#include<cstdio>

using namespace std;

int n;
int g[2001][2001];
int dp[2001][2001];

int Min(int a,int b)
{
    return a<b?a:b;
}

int f(int x,int y)
{
    if(dp[x][y]!=-1)
        return dp[x][y];
    if(g[x][y]==0)
        return dp[x][y]=0;
    if(f(x-1,y)==f(x,y-1))
        if(dp[x-1][y]==0)
            return dp[x][y]=1;
        else
        {
            if(g[x-dp[x-1][y]][y-dp[x-1][y]]==1)
                return dp[x][y]=dp[x-1][y]+1;
            else
                return dp[x][y]=dp[x-1][y];
        }
    else
        if(dp[x][y-1]==0||dp[x-1][y]==0)
            return dp[x][y]=1;
        else
            return dp[x][y]=Min(dp[x-1][y],dp[x][y-1])+1;
}


int main()
{
    int t;

    scanf("%d",&t);

    while(t--)
    {
        char s[2050];
        scanf("%d",&n);
        getchar();

        memset(dp,-1,sizeof(dp));
        for(int i=0;i<=n;i++)
            dp[i][0]=0;
        for(int j=0;j<=n;j++)
            dp[0][j]=0;

        for(int i=1;i<=n;i++)
        {
            gets(s);
            for(int j=1;j<=n;j++)
                g[i][j]=s[j-1]-'0';
        }
        for(int i=1;i<=n;i++)
            for(int j=1;j<=n;j++)
                f(i,j);
        int ans=0;
        for(int i=1;i<=n;i++)
            for(int j=1;j<=n;j++)
                if(dp[i][j]>ans)
                    ans=dp[i][j];
        printf("%d
",ans);
    }

    return 0;
}