Uva--10827（贪心）-优快云博客

2014-07-27 14:31:33

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

思路：通过把矩阵扩展成四倍，然后就转化成二维的求最大连续和的问题了，（注意最长的长度不能超过n，所以要枚举起点把长度限制一下）。

 1 /*************************************************************************
 2     > File Name: p.cpp
 3     > Author: Nature
 4     > Mail: 564374850@qq.com
 5     > Created Time: Sat 26 Jul 2014 01:46:52 PM CST
 6 ************************************************************************/
 7 
 8 #include <cstdio>
 9 #include <cstring>
10 #include <cstdlib>
11 #include <cmath>
12 #include <iostream>
13 #include <algorithm>
14 using namespace std;
15 const int INF = 0x3f3f3f3f;
16 
17 int g[400][400],sumc[400][400];
18 
19 int main(){
20     int Case,n,m;
21     scanf("%d",&Case);
22     while(Case--){
23         scanf("%d",&n);
24         m = 2 * n;
25         for(int i = 1; i <= n; ++i){
26             for(int j = 1; j <= n; ++j){
27                 scanf("%d",&g[i][j]);
28                 g[i][j + n] = g[i][j];
29                 g[i + n][j] = g[i][j];
30                 g[i + n][j + n] = g[i][j];
31             }
32         }
33         for(int i = 1; i <= m; ++i)
34             for(int j = 1; j <= m; ++j)
35                 sumc[i][j] = sumc[i - 1][j] + g[i][j];
36         int ans = -INF;
37         for(int i = 1; i <= m; ++i){
38             for(int j = i; j <= m && j <= i + n - 1; ++j){
39                 for(int v = 1; v <= n; ++v){
40                     int tsum = 0,tmax = -INF;
41                     for(int k = v; k <= v + n - 1; ++k){
42                         tsum += sumc[j][k] - sumc[i - 1][k];
43                         if(tsum < 0 && k < v + n - 1){
44                             tsum = sumc[j][k + 1] - sumc[i - 1][k + 1];
45                             ++k;
46                         }
47                         if(tsum > tmax)
48                             tmax = tsum;
49                     }
50                     ans = tmax > ans ? tmax : ans;
51                 }
52             }
53         }
54         printf("%d\n",ans);
55     }
56     return 0;
57 }