学步园

Maximum sum on a torus
Input: Standard Input

Output: Standard Output

A grid that wraps both horizontally and vertically is called a torus. Given a torus where each cell contains an integer, determine the sub-rectangle with the largest sum. The sum of a sub-rectangle is the sum of all the elements in that rectangle. The grid
below shows a torus where the maximum sub-rectangle has been shaded.

Input

The first line in the input contains the number of test cases (at most 18). Each case starts with an integer N (1≤N≤75) specifying the size of the torus (always square). Then follows N lines describing the torus, each line containing N integers between-100
and 100, inclusive.

Output

For each test case, output a line containing a single integer: the maximum sum of a sub-rectangle within the torus.

Sample Input                                 

2

5

1 -1 0 0 -4

2 3 -2 -3 2

4 1 -1 5 0

3 -2 1 -3 2

-3 2 4 1 -4

3

1 2 3

4 5 6

7 8 9

Output for Sample Input

15
45
http://uva.onlinejudge.org/index.php?option=com_onlinejudge&Itemid=8&category=24&page=show_problem&problem=1768
这道题是求最大子矩阵和的加强版，原矩阵是一个上下左右都可以串通的矩阵环，通过在原矩阵的下方，右方和右下方添加相同的矩阵，然后枚举所求最大子矩阵在第一个矩阵中的左上角，再通过动态规划的方法求出长宽不大于N的最大子矩阵，各种枚举情况中的最大和即为所求解。
#include <cstdio>
using namespace std;

#define MAX 155

int main()
{
	int x,y,i,j,n,s,ans;
	int c[MAX],b[MAX],a[MAX][MAX];
	scanf("%d",&s);
	while (--s>=0){
		scanf("%d",&n);
		for(i=0;i<n;++i)
			for (j=0;j<n;++j){
				scanf("%d",&a[i][j]);
				a[i][j+n]=a[i+n][j]=a[i+n][j+n]=a[i][j];
			}
		ans=-200;
		for (x=0;x<n;++x)
			for(y=0;y<n;++y)
				for(i=0;i<n;++i)
					for(j=0;j<n;++j)
					{
						c[j]=a[x+i][y+j];
						if(j)
							c[j]+=c[j-1];
						if(i)
							b[j]+=c[j];
						else
							b[j]=c[j];
						if (b[j]>ans)
							ans=b[j];
					}
		printf("%d\n",ans);
	}
	return 0;
}