codeforces B. Shower Line 解题报告

题目链接：http://codeforces.com/contest/431/problem/B

题目意思：给出5 * 5 的矩阵。从这个矩阵中选出合理的安排次序，使得happiness之和最大。当第i个人和第j个人talk 的时候，第i个人获得的happiness是g[i][j]，第j 个人获得的happiness是g[j][i]。

好简单的一道题目，不知道昨晚徘徊好久都不敢打，五个for循环即可！数据量这么小......今天一次就过了...

谨以此来纪念自己的怯懦....

方法一：直接暴力枚举

 #include <iostream>
 #include <cstdio>
 #include <cstdlib>
 #include <cstring>
 using namespace std;

 #define LL long long
 int g[][];
 LL ans;

 int main()
 {
     for (int i = ; i <= ; i++)
     {
         for (int j = ; j <= ; j++)
             scanf("%d", &g[i][j]);
     }
     ans = ;
     for (int i = ; i <= ; i++)
     {
         for (int j = ; j <= ; j++)
         {
             if (i != j)
             {
                 for (int k = ; k <= ; k++)
                 {
                     if (k != j && k != i)
                     {
                         for (int l = ; l <= ; l++)
                         {
                             if (l != k && l != i && l != j)
                             {
                                 for (int p = ; p <= ; p++)
                                 {
                                     if (p != l && p != i && p != j && p != k)
                                     {
                                  //       printf("i = %d, j = %d, k = %d, l = %d, p = %d\n", i, j, k, l, p);
                                         LL sum = g[i][j] + g[j][i] + g[j][k] + g[k][j] +  * (g[l][p] + g[p][l] + g[k][l] + g[l][k]);
                                         ans = max(ans, sum);
                                     }
                                 }
                             }
                         }
                     }
                 }
             }
         }
     }
     printf("%lld\n", ans);
     return ;
 }

方法二：利用next_permutation （学人家代码写的）

 #include <iostream>
 #include <cstdio>
 #include <cstdlib>
 #include <algorithm>
 using namespace std;

 #define LL long long
 const int maxn = ;
 int grid[maxn+][maxn+];
 int t[maxn];
 LL ans, tmp;

 int main()
 {
     for (int i = ; i < maxn; i++)
     {
         for (int j = ; j < maxn; j++)
             scanf("%d", &grid[i][j]);
     }
     for (int i = ; i < maxn; i++)
         t[i] = i;
     ans = ;
     do
     {
         //0123: 01 talk  23 talk
         tmp = grid[t[]][t[]] + grid[t[]][t[]];
         tmp += grid[t[]][t[]] + grid[t[]][t[]];
         //1234: 12 talk  34 talk
         tmp += grid[t[]][t[]] + grid[t[]][t[]];
         tmp += grid[t[]][t[]] + grid[t[]][t[]];
         //23: 23 talk
         tmp += grid[t[]][t[]] + grid[t[]][t[]];
         //34: 34 talk
         tmp += grid[t[]][t[]] + grid[t[]][t[]];

         ans = max(ans, tmp);
     }while (next_permutation(t, t+maxn));
     printf("%lld\n", ans);
     return ;
 }