hdu_5691_Sitting in Line(狀壓DP)

題目連接：http://acm.hdu.edu.cn/showproblem.php?pid=5691

題意：中文，不解釋

題解：設dp[i][j]表示當前狀態爲i，以第j個數爲末尾的最憂解，然後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);//返回該數的二進制1的個數
			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;
}