Marriage Ceremonies LightOJ - 1011

最新推荐文章于 2021-05-30 17:06:42 发布

转载最新推荐文章于 2021-05-30 17:06:42 发布 · 104 阅读

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原文链接：http://www.cnblogs.com/zgglj-com/p/7502373.html

本文介绍了一种基于动态规划解决最优婚姻匹配问题的方法，通过计算每对男女之间的优先级指数，实现最大化整体匹配质量的目标。该算法适用于需要进行大量配对且追求最高效率的情景。

You work in a company which organizes marriages. Marriages are not that easy to be made, so, the job is quite hard for you.

The job gets more difficult when people come here and give their bio-data with their preference about opposite gender. Some give priorities to family background, some give priorities to education, etc.

Now your company is in a danger and you want to save your company from this financial crisis by arranging as much marriages as possible. So, you collect N bio-data of men and N bio-data of women. After analyzing quite a lot you calculated the priority index of each pair of men and women.

Finally you want to arrange N marriage ceremonies, such that the total priority index is maximized. Remember that each man should be paired with a woman and only monogamous families should be formed.

Input

Input starts with an integer T (≤ 100), denoting the number of test cases.

Each case contains an integer N (1 ≤ n ≤ 16), denoting the number of men or women. Each of the next N lines will contain N integers each. The j^th integer in the i^th line denotes the priority index between the i^th man and j^th woman. All the integers will be positive and not greater than 10000.

Output

For each case, print the case number and the maximum possible priority index after all the marriages have been arranged.

Sample Input

1 5

2 1

1 2 3

6 5 4

8 1 2

Sample Output

Case 1: 7

Case 2: 16

题解：dp[ i ][ j ]表示第 i 行，状态为 j 时所取得的最大值

 1 #include<cstdio>
 2 #include<cstring>
 3 #include<iostream>
 4 #include<algorithm>
 5 using namespace std;
 6 
 7 int n;
 8 int map[20][20],dp[17][1<<17];
 9 
10 int solve(){
11     for(int i=1;i<=n;i++){
12         for(int j=1;j<(1<<n);j++){
13             int cnt=0;                                 //  ①               
14             for(int k=0;k<n;k++) if(j&(1<<k)) cnt++;   //  ②这两行可写成 int cnt=__builtin_popcount(j)，快了200ms
15             if(cnt!=i) continue;                       //  保证状态中有 i 个1
16             for(int k=0;k<n;k++)
17                 if(j&(1<<k)) dp[i][j]=max(dp[i][j],dp[i-1][j^(1<<k)]+map[i][k+1]);
18         }
19     }
20     return dp[n][(1<<n)-1];
21 }
22 
23 int main()
24 {   int kase;
25     cin>>kase;
26     for(int t=1;t<=kase;t++){
27         cin>>n;
28         for(int i=1;i<=n;i++)
29             for(int j=1;j<=n;j++)
30                 scanf("%d",&map[i][j]);
31         memset(dp,0,sizeof(dp));
32         printf("Case %d: %d\n",t,solve());
33     }
34     return 0;
35 }

转载于:https://www.cnblogs.com/zgglj-com/p/7502373.html