Lightoj 1011 - Marriage Ceremonies

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	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: 1

n男  n女 选n对使他们的权值最大。相当于每行取一个数且这些数不能有在同一列的。

状态：dp[s]=max(dp[s],dp[s|(1<<i)]+a[cnt][i],其中s&(1<<i)=1

/* ***********************************************
Author        :guanjun
Created Time  :2016/6/14 19:21:17
File Name     :1011.cpp
************************************************ */
#include <iostream>
#include <cstring>
#include <cstdlib>
#include <stdio.h>
#include <algorithm>
#include <vector>
#include <queue>
#include <set>
#include <map>
#include <string>
#include <math.h>
#include <stdlib.h>
#include <iomanip>
#include <list>
#include <deque>
#include <stack>
#define ull unsigned long long
#define ll long long
#define mod 90001
#define INF 0x3f3f3f3f
#define maxn 10010
#define cle(a) memset(a,0,sizeof(a))
const ull inf = 1LL << ;
const double eps=1e-;
using namespace std;
priority_queue<int,vector<int>,greater<int> >pq;
struct Node{
    int x,y;
};
struct cmp{
    bool operator()(Node a,Node b){
        if(a.x==b.x) return a.y> b.y;
        return a.x>b.x;
    }
};

bool cmp(int a,int b){
    return a>b;
}
int dp[<<];
int n,m;
int a[][];
int dfs(int s,int cnt){
    if(s==(<<n)-){
        return dp[s]=;
    }
    if(dp[s]!=-)return dp[s];
    for(int i=;i<n;i++){
        if(!(s&(<<i))){
            dp[s]=max(dp[s],dfs(s|(<<i),cnt+)+a[cnt][i]);
        }
    }
    return dp[s];
}
int main()
{
    #ifndef ONLINE_JUDGE
    freopen("in.txt","r",stdin);
    #endif
    //freopen("out.txt","w",stdout);
    int T;
    cin>>T;
    for(int t=;t<=T;t++){
        cin>>n;
        for(int i=;i<n;i++)
            for(int j=;j<n;j++)
                scanf("%d",&a[i][j]);
        memset(dp,-,sizeof dp);
        printf("Case %d: %d\n",t,dfs(,));

    }
    return ;
}

另一种姿势：dp[i][s]=max(dp[i][s],dp[i-1][k]+a[i][j]).第i行状态为s时能得到的最大值。k是上一行的状态。暴力转移一下：

#include <stdio.h>
#include <string.h>
#include <iostream>
using namespace std;
int arr[][];
int dp[][ << ];
int main()
{
    int T;
    int _case = ;
    scanf("%d", &T);
    while(T--)
    {
        memset(dp, , sizeof(dp));
        int n;
        scanf("%d", &n);
        for(int i = ; i <= n; i++)
            for(int j = ; j < n; j++)
                scanf("%d", &arr[i][j]);
        for(int i = ; i <= n; i++)
        {
            for(int j = ; j < n; j++)
            {
                for(int k = ;k <= ( << n) - ;k++)
                {
                    if( (k & ( << j)) == )
                        dp[i][k | ( << j)] = max(dp[i][k | ( << j)], dp[i - ][k] + arr[i][j]);
                }
            }
        }
        int ans = ;
        for(int i = ;i <= ( << n) - ;i++)
            ans = max(ans, dp[n][i]);
        printf("Case %d: %d\n", _case++, ans);
    }
    return ;
}