uva 10817（状压dp)

题意：就是有个学校要招老师．要让没门课至少有两个老师可以上．每个样样例先输入三个数字课程数量ｓ，已经在任的老师数量，和应聘的老师数量．已经在任的一定要聘请．

思路是参考了刘汝佳书上的，关键如何状压。

#include <cstdio>
#include <iostream>
#include <sstream>
#include <cmath>
#include <cstring>
#include <cstdlib>
#include <string>
#include <vector>
#include <map>
#include <set>
#include <queue>
#include <stack>
#include <algorithm>
using namespace std;
#define ll long long
#define _cle(m, a) memset(m, a, sizeof(m))
#define repu(i, a, b) for(int i = a; i < b; i++)
#define repd(i, a, b) for(int i = b; i >= a; i--)
#define sfi(n) scanf("%d", &n)
#define pfi(n) printf("%d\n", n)
#define sfi2(n, m) scanf("%d%d", &n, &m)
#define pfi2(n, m) printf("%d %d\n", n, m)
#define pfi3(a, b, c) printf("%d %d %d\n", a, b, c)
#define MAXN 125
#define MAXS 8
const int INF = 0x3f3f3f3f;
int c[MAXN], m, n, s, st[MAXN], d[MAXN][<<MAXS][<<MAXS];

int dp(int i, int s0, int s1, int s2)
{
    if(i == m + n) return s2 == (<<s) -  ?  : INF;
    int &ans = d[i][s1][s2];
    if(ans >= ) return ans;
    ans = INF;
    if(i >= m) ans = dp(i + , s0, s1, s2);
    int m0 = st[i] & s0, m1 = st[i] & s1;
    s0 ^= m0, s1 = (s1 ^ m1) | m0, s2 |= m1;
    ans = min(ans, c[i] + dp(i + , s0, s1, s2));
    return ans;

}
int main()
{
    while()
    {
        sfi2(s, m), sfi(n);
        if(s == ) break;
        _cle(d, -);
        _cle(st, );
        char ch;
        int ss;
        repu(i, , m + n)
        {
            sfi(c[i]);
            while()
            {
                sfi(ss);
                ch = getchar();
                ss--;
                st[i] += (<<ss);
                if(ch == '\n') break;
            }
        }

        pfi(dp(, (<<s) - , , ));
        //pfi(d[m + n - 1][0][(1<<s) - 1]);
    }
    return ;
}