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最新推荐文章于 2020-04-06 00:53:58 发布

Smile_Benson

最新推荐文章于 2020-04-06 00:53:58 发布

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本文介绍了一个在矩形圆柱面上寻找具有最大和的子矩阵的问题，并提供了一种通过使用前缀和降维计算来提高效率的解决方案。文章详细解释了解题思路并附带了完整的代码实现。

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Maximum sum on a torus

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

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

Sample output

题意: 现在给出一个矩形的圆柱面, 每一个格子都有一个整数, 现在要你求出最大和子矩阵.

解题思路:

1. 求最大子矩阵和, 采用前缀和降维计算可以大大提高效率.

2. 这个问题难点是一般的最大子矩阵变成了再一个圆柱面上计算了, 不过也是没问题. 将原

矩阵扩大2倍即可.

3. ans = max( ans, sum[i+k-1][j+p-1]-sum[i+k-1][j-1]-sum[i-1][j+p-1]+sum[i-1][j-1] );

代码:

#include <cstdio>
#include <iostream>
#include <cstring>
using namespace std;
#define MAX 155
const int INF = (1<<29);

int n, m;
int g[MAX][MAX], sum[MAX][MAX];

inline int max(int a, int b)
{
return a > b ? a : b;
}

int main()
{
// freopen("input.txt", "r", stdin);
int caseNum;
scanf("%d", &caseNum);
while(caseNum--)
{
  scanf("%d", &n);
  int i, j, k, p;
  for(i = 1; i <= n; ++i)
  {
   for(j = 1; j <= n; ++j)
   {
    scanf("%d", &g[i][j]);
    g[i+n][j] = g[i][j];
    g[i][j+n] = g[i][j];
    g[i+n][j+n] = g[i][j];
   }
  }

  memset(sum, 0, sizeof(sum));
  for(i = 1; i <= 2*n; ++i)
   for(j = 1; j <= 2*n; ++j)
    sum[i][j] = sum[i-1][j]+g[i][j];

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

  int ans = -INF;
  for(i = 1; i <= n; ++i)
   for(j = 1; j <= n; ++j)
    for(k = 1; k <= n; ++k)
     for(p = 1; p <= n; ++p)
      ans = max(ans, sum[i+k-1][j+p-1]-sum[i+k-1][j-1]-sum[i-1][j+p-1]+sum[i-1][j-1]);
  printf("%d\n", ans);
}
return 0;
}