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
      
       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 0x3f3f3f3fconst 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<
       
        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<
        
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