lightoj 1011 - Marriage Ceremonies(状态压缩dp) - dblank

lightoj 1011 - Marriage Ceremonies(状态压缩dp)

Time Limit: 2 second(s) Memory Limit: 32 MB

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 jth integer in the ith line denotes the priority index between the ith man and jth 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

Output for Sample Input

2

2

1 5

2 1

3

1 2 3

6 5 4

8 1 2
Case 1: 7

Case 2: 16

 


PROBLEM SETTER: JANE ALAM JAN

题意:给你一个表,第i,j个表示第i个男人和第j个女人匹配可以得到的价值,每个人只能匹配一个异性,求最大价值和。

思路:n的范围只有16,可以用状态压缩,dp[i][j]表示前i个男人匹配状态为j的最大价值和,于是又状态转移 dp[i][j|temp] = max(dp[i][j|temp], dp[i-1][j] + Map[i][k])。

这个题用km明显更快啊。

#include<iostream>  
#include<cstdio>  
#include<algorithm>  
#include<map>  
#include<cstring>  
#include<map>  
#include<set>  
#include<queue>  
#include<stack>  
#include<cmath>  
using namespace std;  
const double eps = 1e-10;  
const int inf = 0x3f3f3f3f, N = 2e1 +10;  
int Map[N][N];  
int dp[N][(1<<16)+10];  
int main()  
{  
  
    int n, t;  
    cin>>t;  
    for(int cas = 1; cas<=t; cas++)  
    {  
        scanf("%d", &n);  
        for(int i = 0; i<n; i++)  
            for(int j = 0; j<n; j++)  
                scanf("%d",&Map[i][j]);  
        memset(dp, 0, sizeof(dp));  
        int temp, res = 1<<n;  
        for(int i = 0; i<n; i++) dp[0][1<<i] = Map[0][i];  
        for(int i = 1; i<n; i++)  
        {  
            for(int j = 0; j<res; j++)  
                for(int k = 0; k<n; k++)  
                {  
                    temp = 1<<k;  
                    if(j & temp)  
                        continue;  
                    dp[i][j|temp] = max(dp[i][j|temp], dp[i-1][j] + Map[i][k]);  
                }  
        }  
        printf("Case %d: %d\n", cas, dp[n-1][res-1]);  
    }  
    return 0;  
}

 

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