Hamburger Magi 状压dp

　　题意: 一个人有m元钱  有n个汉堡  每个汉堡有其价值和其花费   且做某个汉堡需要做好一些前提汉堡    求最大价值

强行状压即可   把两个&写成了&& QAQ 

#include<bits/stdc++.h>
using namespace std;
//input by bxd
#define rep(i,a,b) for(int i=(a);i<=(b);i++)
#define repp(i,a,b) for(int i=(a);i>=(b);--i)
#define RI(n) scanf("%d",&(n))
#define RII(n,m) scanf("%d%d",&n,&m)
#define RIII(n,m,k) scanf("%d%d%d",&n,&m,&k)
#define RS(s) scanf("%s",s);
#define ll long long
#define pb push_back
#define REP(i,N)  for(int i=0;i<(N);i++)
#define CLR(A,v)  memset(A,v,sizeof A)
//////////////////////////////////
#define inf 0x3f3f3f3f
const int N=20;
int n,m;
int cost[N];
int val[N];
int bef[N][N];
int money[N<<15];
int dp[N<<15];

bool judge(int state,int j)
{
    if(state& (1<<(j-1)) )return 0;
    rep(i,1,bef[j][0])
    {
        int x=bef[j][i];
        if( !(state&(1<<(x-1))) )return 0;
    }
    return 1;
}

int main()
{
    int cas;RI(cas);
    while(cas--)
    {
        RII(n,m);
        rep(i,1,n)RI(val[i]);
        rep(i,1,n)RI(cost[i]);
        rep(i,1,n)
        {
            int q;RI(bef[i][0]);
            rep(j,1,bef[i][0])
            RI(bef[i][j]);
        }
        CLR(dp,-0x3f);
        CLR(money,0);
        dp[0]=0;
        int maxx=0;
        rep(i,0, (1<<n)-1 )
        {
            rep(j,1,n)
            if(judge(i,j)&&cost[j]+money[i]<=m&&dp[i]+val[j]>dp[i|(1<<(j-1))] )
            {
                dp[i|(1<<(j-1))]=dp[i]+val[j];
                maxx=max(maxx,dp[i|(1<<(j-1))]);
                money[i|(1<<(j-1))]=cost[j]+money[i];
            }
        }
        cout<<maxx<<endl;
    }
    return 0;
}