題目：

The headmaster of Spring Field School is considering employing some new teachers for certain subjects. There are a number of teachers applying for the posts. Each teacher is able to teach one or more subjects. The headmaster wants to select applicants so that each subject is taught by at least two teachers, and the overall cost is minimized.

Input

The input consists of several test cases. The format of each of them is explained below: The first line contains three positive integers S, M and N. S (≤ 8) is the number of subjects, M (≤ 20) is the number of serving teachers, and N (≤ 100) is the number of applicants. Each of the following M lines describes a serving teacher. It first gives the cost of employing him/her (10000 ≤ C ≤ 50000), followed by a list of subjects that he/she can teach. The subjects are numbered from 1 to S. You must keep on employing all of them. After that there are N lines, giving the details of the applicants in the same format. Input is terminated by a null case where S = 0. This case should not be processed.

Output

For each test case, give the minimum cost to employ the teachers under the constraints.

Sample Input

2 2 2

10000 1

20000 2

30000 1 2

40000 1 2

0 0 0

Sample Output

60000

意思：

某校有m個教師和n個求職者，需要教授s個課程（1<=s<=8,1<=m<=20,1<=n<=100）.已知每人的工資c(10000<=c<=50000)和能教的課程集合，要求支付最少的工資使每門課都至少有兩名教師能教。在職教師不能辭退。

之後補充做法

#include<cstdio>
#include<iostream>
#include<cstring> 
#include<algorithm>
#include<sstream>
#define maxn 150  
#define INF 0X3F3F3F3F  
using namespace std;  
  
int S,M,N;  
int st[maxn];  
int c[maxn];  
char str[100];  
int d[maxn][1<<8][1<<8];  
  
int dp(int i,int s0,int s1,int s2)  
{  
    if(i==M+N) return s2==(1<<S)-1?0:INF;  
    int& ans=d[i][s1][s2];  
    if(ans>=0) return ans;  
    ans=INF;  
    if(i>=M) ans=dp(i+1,s0,s1,s2);  
    int m0=st[i]&s0;  
    int m1=st[i]&s1;  
    s0^=m0;  
    s1^=m1;  
    s1|=m0;  
    s2|=m1;  
    ans=min(ans,dp(i+1,s0,s1,s2)+c[i]);  
    return ans;  
}  
  
int main()  
{  
    while(scanf("%d %d %d",&S,&M,&N),S||M||N)  
    {  
        getchar();  
        memset(d,-1,sizeof(d));  
        memset(st,0,sizeof(st));  
        string s;  
        for(int i=0;i<M+N;i++)  
        {  
            cin.getline(str,100);  
            s=str;  
            stringstream ss(s);  
            int temp;  
            ss>>c[i];  
            while(ss>>temp) st[i]|=1<<(temp-1);  
        }  
        printf("%d\n",dp(0,(1<<S)-1,0,0));  
    }  
    return 0;  
}  

Headmaster's Headache UVa10817【DP】（缺）