1011 - Marriage Ceremonies

http://lightoj.com/volume_showproblem.php?problem=1011

状压dp

1011 - Marriage Ceremonies

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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 j^thinteger 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: 16

 

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PROBLEM SETTER: JANE ALAM JAN

第一道状压dp，感觉状压dp的特点就是n特别小，最多就只能到20，再大就要爆了。

设dp[i][j]表示前i行状态为j的最大值，这一题跟放花那一题有什么区别呢？为什么不能像放花一样解，原因在与放花那一题有前i朵花怎么放这么一个子问题，这一题却没有，因为没有顺序。

#include <cstdio>#include <cstring>#include <algorithm>#include <vector>#define LL long longusing namespace std;const int maxn=1<<30;const int SIZE=1e2+10;int cmap[20][20],dp[17][70000];int main(){    int T,n;    scanf("%d",&T);    for(int cas=1;cas<=T;cas++){        scanf("%d",&n);        for(int i=1;i<=n;i++)            for(int j=1;j<=n;j++)                scanf("%d",&cmap[i][j]);        memset(dp,0,sizeof(dp));        for(int i=1;i<=n;i++){//第i行枚举前i-1行状态j            for(int j=0;j<(1<<n);j++){                int cnt=0;                for(int k=0;k<n;k++){                    if(j&(1<<k))cnt++;                }                if(cnt!=i-1)continue;                for(int k=0;k<n;k++){                    if(j&(1<<k))continue;                    dp[i][j|(1<<k)]=max(dp[i][j|(1<<k)],dp[i-1][j]+cmap[i][k+1]);//注意是第k+1个位置可以放                }            }        }        printf("Case %d: %d\n",cas,dp[n][(1<<n)-1]);    }    return 0;}