hdu_3182_Hamburger Magi(状压DP)

题目连接:http://acm.hdu.edu.cn/showproblem.php?pid=3182

题意：有n个汉堡，做每个汉堡需要消耗一定的能量，每个汉堡对应一定的价值，且只能做一次，并且做当前汉堡需要先做出列出的汉堡，求最大的价值

题解：状压DP

#include<cstdio>
#define FFC(i,a,b) for(int i=a;i<=b;++i)

int T,n,all,end,now,cur,ans,flag,v[16],e[16],a[16][16],dp[1<<16],c[1<<16];

int main(){
    scanf("%d",&T);
    while(T--){
        scanf("%d%d",&n,&all),end=(1<<n)-1,c[0]=all;
        FFC(i,1,n)scanf("%d",&v[i]);FFC(i,1,n)scanf("%d",&e[i]);
        FFC(i,1,n){
            scanf("%d",&a[i][0]);
            FFC(j,1,a[i][0])scanf("%d",&a[i][j]);
        }
        FFC(i,0,end)dp[i]=-1;
        dp[0]=0,ans=0;
        FFC(i,0,end){
            if(dp[i]==-1)continue;
            FFC(j,0,n-1){
                now=1<<j,flag=0,cur=now|i;
                if((i&now)==0&&c[i]>=e[j+1]){
                FFC(k,1,a[j+1][0])if(((1<<(a[j+1][k]-1))&i)==0){flag=1;break;}
                if(flag)continue;
                if(dp[cur]<dp[i]+v[j+1])
                dp[cur]=dp[i]+v[j+1],c[cur]=c[i]-e[j+1],ans=ans>dp[cur]?ans:dp[cur];
                }
            }
        }
    printf("%d\n",ans);
    }
    return 0;
}