ACM: uva 10827 -&…

Maximum sum on a torus

A grid that wraps both horizontally and vertically is called atorus. Given a torus where each cell contains an integer, determinethe sub-rectangle with the largest sum. The sum of a sub-rectangleis the sum of all the elements in that rectangle. The grid belowshows 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 Nlines 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

 

Sample output

15

45

 

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;
}