Hamburger Magi 状压dp

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;
}

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