Marriage Ceremonies(状态压缩dp）

 Marriage Ceremonies

Time Limit:2000MS     Memory Limit:32768KB     64bit IO Format:%lld & %llu

Submit Status Practice LightOJ 1011

Description

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^thwoman. 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

2

2

1 5

2 1

3

1 2 3

6 5 4

8 1 2

Sample Output

Case 1: 7

Case 2: 16

 

#include<stdio.h>
#include<string.h>
#include<algorithm>
using namespace std;
int a[17][17] ;
int dp[17][1 << 17] ;
int n ;

int main ()
{
   // freopen ("a.txt" , "r" , stdin ) ;
    int T ;
    scanf ("%d" , &T ) ;
    int ans = 0 ;
    while (T--) {
        scanf ("%d" , &n ) ;
        for (int i = 0 ; i < n ; i++) {
            for (int j = 0 ; j < n ; j++) {
                scanf ("%d" , &a[i][j]) ;
            }
        }
        for (int i = 0 ; i < n ; i++) {
            for (int j = 0 ; j < 1 << n ; j++ ) {
                dp[i][j] = -1 ;
            }
        }
        for (int i = 0 ; i < n ; i++) {
            dp[0][1 << i] = a[0][i] ;
        }
        for (int i = 1 ; i < n ; i++) {
            for (int j = 0 ; j < 1 << n ; j++) {
                if (dp[i - 1][j] != -1) {
                    for (int k = 0 ; k < n ; k++) {
                        if ( ! (j & 1 << k) ) {
                            dp[i][j | 1 << k] = max (dp[i][j | 1 << k] , dp[i - 1][j] + a[i][k] ) ;
                        }
                    }
                }
            }
        }
        printf ("Case %d: %d
" , ++ans , dp[n - 1][(1 << n) - 1 ] ) ;
    }
    return 0 ;
}

先把“ 数据 ”存入 下标为 1 << 0 , 1 << 1 , 1 << 2 , 1 << n - 1的这几个位置，然后就是一般的dp过程了

状态压缩dp我觉得有两点蛮神奇的（当然能用数学方法证明的）
：

1，! ( j & 1 << k ) 这能让已经加过的滚蛋；orz

2，(1 << 0) + (1 << 1 ) + ... + (1 << n - 1) = (1 << n ) - 1 orz 不仔细看还真没发觉