hdu_5691_Sitting in Line(状压DP)

题目连接：http://acm.hdu.edu.cn/showproblem.php?pid=5691

题意：中文，不解释

题解：设dp[i][j]表示当前状态为i，以第j个数为末尾的最忧解，然后dp下去就行了

 

#include<cstdio>
#define F(i,a,b) for(int i=a;i<=b;i++)
inline void up(int &x,int y){if(x<y)x=y;}

int t,n,dp[1<<16][17],a[17],b[17],inf=-(1<<30),end,ans,ic=1;

int main(){
    scanf("%d",&t);
    while(t--){
        scanf("%d",&n),end=(1<<n)-1,ans=inf;
        F(i,1,n)scanf("%d%d",a+i,b+i),b[i]++;
        F(i,0,end)F(j,1,n)dp[i][j]=inf;
        F(i,1,n)if(!b[i]||b[i]==1)dp[1<<(i-1)][i]=0;
        F(i,0,end){
            int have=__builtin_popcount(i);//返回该数的二进制1的个数
            F(j,1,n)if(dp[i][j]>inf)
            F(k,1,n)if(j==k||(b[k]&&b[k]!=have+1)||i&1<<(k-1))continue;
            else up(dp[i|1<<(k-1)][k],dp[i][j]+a[j]*a[k]);
        }
        F(i,1,n)up(ans,dp[end][i]);
        printf("Case #%d:\n%d\n",ic++,ans);
    }
    return 0;
}