king's trouble II

最新推荐文章于 2021-11-22 13:11:26 发布

wanghandou

最新推荐文章于 2021-11-22 13:11:26 发布

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分类专栏：动态规划

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本文介绍了一种使用动态规划解决矩阵中寻找最大全为1的正方形子矩阵问题的方法。通过构建一个二维数组来记录每个位置能形成的正方形的最大边长，最终找到整个矩阵中的最大值。

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Description
Long time ago, a king occupied a vast territory.
Now there is a problem that he worried that he want to choose a largest square of his territory to build a palace.
Can you help him?
For simplicity, we use a matrix to represent the territory as follows:
0 0 0 0 0
0 1 0 1 0
1 1 1 1 0
0 1 1 0 0
0 0 1 0 0
Every single number in the matrix represents a piece of land, which is a 1*1 square
1 represents that this land has been occupied
0 represents not
Obviously the side-length of the largest square is 2

Input
The first line of the input contains a single integer t (1≤t≤5) — the number of cases.
For each case
The first line has two integers N and M representing the length and width of the matrix
Then M lines follows to describe the matrix
1≤N,M≤1000

Output
For each case output the the side-length of the largest square

Sample Input
2
5 5
0 0 0 0 0
0 1 0 1 0
1 1 0 1 0
0 1 1 0 0
0 0 1 0 0
5 5
0 0 0 0 0
0 1 0 1 0
1 1 1 1 0
0 1 1 0 0
0 0 1 0 0

Sample Output
1
2

#include<stdio.h>
#include<iostream>
#include<string.h>
using namespace std;
int a[1010][1010];
int dp[1010][1010];
int main()
{
    int t;
    scanf("%d",&t);
    while(t--)
    {
        int n,m;
        scanf("%d%d",&n,&m);
        memset(dp,0,sizeof(dp));
        for(int i=0;i<n;i++)
        {
            for(int j=0;j<m;j++)
            {
                scanf("%d",&a[i][j]);
            }
        }
        int ans=0;
        for(int i=1;i<n;i++)
        {
            for(int j=1;j<m;j++)
            {
                if(a[i][j]==1)
                {
                    dp[i][j]=min(dp[i][j-1],min(dp[i-1][j-1],dp[i-1][j]))+1;
                }
                 ans=max(ans,dp[i][j]);
            }
        }
        printf("%d\n",ans);
    }
}