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CSU 1424 Qz’s Maximum All One Square

2024-07-17 05:20:35   215人阅读

Description

     YH gave Qz an odd matrix consists of one or zero. He asked Qz to find a square that has the largest area. The problem is not so easy, that means the square you find must not contain any zero. Your task is finding the square and you only need to output the area of the matrix. If you help Qz, he will thanks you a lot and treat you a meal. Please call him after the contest if you solve this problem. His number is XXXXXXXXXXXX. Hehe~

Input

For each case, Qz first give you 2 number n and m (1 <= n <= 1000, 1 <= m <= 1000),the matrix is in size of n columns and m rows.
Then Qz will give you the matrix. To avoid cost his money, Qz decided to make this problem a little difficult... Do not worry, he just gave you multiple cases.

Output

        For each test case, output the desired answer.

Sample Input

2 2
1 1
1 1
4 4
1 1 0 0
1 1 1 1
1 1 1 0
1 1 0 1

Sample Output

4
4


这是道很老的题，各大题库基本都有，以前也做过类似的题目。可能是这几天为了私事花了很多心思，做题的时候总是不在状态。

DP应该是一眼就能看出来的，状态转移方程也很简单，需要注意的一点是行号和列号的标示，尤其是在还没有十分清晰明确的思路时，我经常写着写着就搞混了。

#include<stdio.h>
#define maxn 1010

int sum[maxn][maxn];
int main()
{
    int m,n,t;
    while(~scanf("%d%d",&n,&m))
    {
        int i,j,k,ans;
        for(i = 1;i <= m;i++)
        {
            for(j = 1;j <= n;j++)
            {
                scanf("%d",&k);
                sum[i][j] = k + sum[i-1][j] + sum[i][j-1] - sum[i-1][j-1];//求(i,j）到（1，1）这个点的1的个数。
            }
        }
        ans = 0;
        for(i = 1;i <= m;i++)
            for(j = 1;j <= n;j++)
            {
                t = ans;//每次从边长等于ans开始找（剪枝，）
                int x,y;
                for(x= i - ans,y = j - ans ; x && y ; x--,y--)
                {
                    if(x <= 0 || y <= 0) break;
                    int temp1 = sum[i][j] - sum[i][y-1] - sum[x-1][j] + sum[x-1][y-1];
                    if(temp1 == (t + 1) * (t + 1))
                    {
                        t++;
                    }
                    else break;
                }
                if(t > ans) ans = t;
            }
        printf("%d\n",ans * ans);
    }
    return 0;
}






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