UVa 10827

　　题目大意：UVa 108 - Maximum Sum的加强版，求最大子矩阵和，不过矩阵是可以循环的，矩阵到结尾时可以循环到开头。开始听纠结的，想着难道要分情况讨论吗？！就去网上搜，看到可以通过补全进行处理，也是，通过补全一个相同的，问题就迎刃而解了，所以把n*n的矩阵扩展成2n*2n的矩阵就好了。

#include <cstdio>
#include <cstring>
#define MAXN 160

int a[MAXN][MAXN], sum[MAXN][MAXN];

int main()
{
#ifdef LOCAL
    freopen("in", "r", stdin);
#endif
    int T;
    scanf("%d", &T);
    while (T--)
    {
        int n;
        scanf("%d", &n);
        memset(sum, 0, sizeof(sum));
        for (int i = 1; i <= n; i++)
        {
            for (int j = 1; j <= n; j++)
                scanf("%d", &a[i][j]);
            for (int j = n+1; j <= 2*n; j++)
                a[i][j] = a[i][j-n];
        }
        for (int i = n+1; i <= 2*n; i++)
            for (int j = 1; j <= 2*n; j++)
                a[i][j] = a[i-n][j];
        for (int i = 1; i <= 2*n; i++)
            for (int j = 1; j <= 2*n; j++)
                sum[i][j] = sum[i][j-1] + sum[i-1][j] - sum[i-1][j-1] + a[i][j];
        int max = a[1][1];
        for (int i = 1; i <= n; i++)
            for (int p = 0; p < n; p++)
                for (int j = 1; j <= n; j++)
                    for (int q = 0; q < n; q++)
                    {
                        int t = sum[i+p][j+q] - sum[i+p][j-1] - sum[i-1][j+q] + sum[i-1][j-1];
                        if (t > max)   max = t;
                    }
        printf("%d
", max);
    }
    return 0;
}