Maximum sum on a torus UVA-优快云博客

本文链接：https://blog.youkuaiyun.com/abc123456789cbaabc/article/details/8655532

本文介绍如何在给定的二维网格上找到具有最大和的矩形子集，涉及动态规划和前缀和技巧。

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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.

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

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 Output for Sample Input

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

1 2 3

4 5 6

7 8 9

也可参考：http://blog.youkuaiyun.com/vinci_lau/article/details/7078968（利用动态规划解决的问题）

以下是自己利用前缀和解决的问题；

关键是：

因为torus存在一种“可包围循环wrap”的情况

所以必须把给定的NXN的矩阵变成2NX2N的矩阵

然后利用前缀和，在枚举的时候注意枚举的矩阵必须包含在

NXN的矩阵中（即行列不超过N）

代码如下

max(0,j-n)//防止j-n<0

#include <stdio.h>
#include <math.h>
#include <algorithm>
#include <string.h>
#include <stdlib.h>
using namespace std;
int grid[160][160];
int sum[160][160];
int main()
{
    int t;
    scanf("%d",&t);
    for(int i = 0;i<t;++i)
    {
        int n;
        scanf("%d",&n);
        for(int j = 1;j<=n;++j)
          for(int k = 1;k<=n;++k)
             {
                 scanf("%d",&grid[j][k]);
                 grid[j+n][k] = grid[j][k];
                 grid[j][k+n] = grid[j][k];
                 grid[j+n][k+n] = grid[j][k];
             }
        for(int j = 0;j<=2*n;++j)
        {
            sum[0][j] = 0;
            sum[j][0] = 0;
        }

        for(int j = 1;j<=2*n;++j)
            for(int k = 1;k<=2*n;++k)
                sum[j][k] = sum[j][k-1] + sum[j-1][k] - sum[j-1][k-1] + grid[j][k];

        int maxn = grid[1][1];
       for(int j = 2*n;j>=1;--j)
         for(int k = 2*n;k>=1;--k)
           for(int a =max(0,j-n) ;a<j;++a)
             for(int b =max(k-n,0);b<k;++b)
             {
                 int temp = sum[j][k] - sum[j][b] - sum[a][k] + sum[a][b];
                 maxn = max(maxn,temp);
             }

          printf("%d\n",maxn);
    }
    return 0;
}