题意：某个学校有m个老师和n个求职者，需要讲授s个课程，已知每个人的工资c和能交的课程，求花费最小使得每门课程都至少有两个人教。

思路：状压dp，将每个老师要交的课程压缩成一个数，然后对于每门课，找到每个老师教与不交的最小状态即可。（因为INF还被坑了几次？？）

code：

#include <iostream>
#include <cstdio>
#include <cmath>
#include <algorithm>
#include <cstring>
#include <sstream>
#include <string>
#include <vector>
#include <list>
#include <queue>
#include <stack>
#include <map>
#include <set>
#include <bitset>using namespace std;typedef long long ll;
typedef unsigned long long ull;
typedef long double ld;const int INF=9999999;
const int inf=-INF;
const int N=125;
const int M=2005;
const int mod=1000000007;
const double pi=acos(-1.0);#define cls(x,c) memset(x,c,sizeof(x))
#define ft(i,s,n) for (int i=s;i<=n;i++)
#define lson l,m,rt<<1
#define rson m+1,r,rt<<1|1
#define lrt  rt<<1
#define rrt  rt<<1|1
#define middle int m=(r+l)>>1
#define lowbit(x) (x&-x)
#define pii pair<int,int>
#define mk make_pair
#define IN freopen("in.txt","r",stdin);
#define OUT freopen("out.txt","w",stdout);int n,m,s,c[N],st[N],d[N][1<<8][1<<8];int dp(int i,int s0,int s1,int s2)
{if (i==m+n) { if (s2==((1<<s)-1)) return 0; return INF;}int& ans=d[i][s1][s2];if (ans>=0) return ans;ans=INF;if (i>=m) ans=min(ans,dp(i+1,s0,s1,s2));int m=st[i]&s0,m1=st[i]&s1;//s0^=m0; s1=(s1^m1)|m0;s2|=m1;ans=min(ans,c[i]+dp(i+1,s0^m,(s1^m1)|m,s2|m1));return ans;
}
int main()
{while (~scanf("%d %d %d",&s,&m,&n),s){ft(i,0,n+m-1){scanf("%d",&c[i]);st[i]=0;char ch=getchar();while (ch!='\n'){if (ch>='1'&&ch<='9') {int t=ch-'1';st[i]=(st[i]|(1<<t));}ch=getchar();}//cout<<st[i]<<endl;}cls(d,-1);printf("%d\n",dp(0,(1<<s)-1,0,0));}
}