CSU 1424 Qz’s Maximum All One Square

　　原题链接：http://acm.csu.edu.cn/OnlineJudge/problem.php?id=1424

　　逐渐找到做这种题的感觉了。

　　二分法。g[i][j]存储坐标(i, j)的值，s[i][j]存储的值为左上角为起始点(1,1)，右下角为(i,
j)的矩形区域内所有值的和，那么：

　　　　s[i][j] = g[i][j] + s[i-1][j] + s[i][j-1] - s[i-1][j-1]

　　扫描整个矩形，遇到为“1”的点就将其作为起点，开始二分边长，利用数组s在O(1)的时间复杂度内判断是否满足为由1组成的正方形，不管更新最大值即可。

#include <stdio.h>

#include <string.h>

#include <algorithm>

using namespace std;

#define N 1005
int g[N][N], s[N][N];
int n, m;
bool ok(int i, int j, int mid)

{

    int t1 = mid * mid;

    int t2 = s[i+mid-1][j+mid-1] - s[i+mid-1][j-1] - s[i-1][j+mid-1] + s[i-1][j-1];

    return t1 == t2;

}
int bs(int i, int j)

{

    int l = 1, r = N;

    while(l < r)

    {

        int mid = (l + r) >> 1;

        if(i + mid - 1 > m || j + mid - 1 > n)

            r = mid;

        else if(ok(i, j, mid))

            l = mid+1;

        else

            r = mid;

    }

    return (l-1) * (l-1);

}
int main()

{

    int ans;

    while(scanf("%d%d", &n, &m) != EOF)

    {

        for(int i = 1; i <= m; i++)

            for(int j = 1; j <= n; j++)

                scanf("%d", &g[i][j]);
for(int i = 1; i <= m; i++)

            for(int j = 1; j <= n; j++)

                s[i][j] = g[i][j] + s[i-1][j] + s[i][j-1] - s[i-1][j-1];

        ans = 0;

        for(int i = 1; i <= m; i++)

            for(int j = 1; j <= n; j++)

                if(g[i][j])

                    ans = max(ans, bs(i, j));

        printf("%d\n", ans);

    }

    return 0;

}

CSU 1424 Qz’s Maximum All One Square,布布扣,bubuko.com