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

 1 #include<cstdio>
 2 #define F(i,a,b) for(int i=a;i<=b;i++)
 3 inline void up(int &x,int y){if(x<y)x=y;}
 4
 5 int t,n,dp[1<<16][17],a[17],b[17],inf=-(1<<30),end,ans,ic=1;
 6
 7 int main(){
 8     scanf("%d",&t);
 9     while(t--){
10         scanf("%d",&n),end=(1<<n)-1,ans=inf;
11         F(i,1,n)scanf("%d%d",a+i,b+i),b[i]++;
12         F(i,0,end)F(j,1,n)dp[i][j]=inf;
13         F(i,1,n)if(!b[i]||b[i]==1)dp[1<<(i-1)][i]=0;
14         F(i,0,end){
15             int have=__builtin_popcount(i);//返回该数的二进制1的个数
16             F(j,1,n)if(dp[i][j]>inf)
17             F(k,1,n)if(j==k||(b[k]&&b[k]!=have+1)||i&1<<(k-1))continue;
18             else up(dp[i|1<<(k-1)][k],dp[i][j]+a[j]*a[k]);
19         }
20         F(i,1,n)up(ans,dp[end][i]);
21         printf("Case #%d:\n%d\n",ic++,ans);
22     }
23     return 0;
24 }

View Code

转载于:https://www.cnblogs.com/bin-gege/p/5696102.html

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