hdu_5418_Victor and World(状压DP+Floyd)

题目连接:http://acm.hdu.edu.cn/showproblem.php?pid=5418

题意:给你n个点,和一些边,找一条路径经过全部的点,并回到起点,问最小的花费是多少,

题解:m<=1e5,n<16,说明有多边,需要处理一下,处理完用floyd跑一下,然后进行状态压缩dp。


 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 dp[1<<16][17],g[17][17],inf=1<<29,n,m,t,u,v,c,end,ans;
 6 
 7 int main(){
 8     scanf("%d",&t);
 9     while(t--){
10         scanf("%d%d",&n,&m),end=(1<<n)-1,ans=inf;
11         F(i,1,n)F(j,1,n)g[i][j]=(i==j?0:inf);
12         F(i,1,m)scanf("%d%d%d",&u,&v,&c),up(g[u][v],c),up(g[v][u],c);
13         F(k,1,n)F(i,1,n)F(j,1,n)up(g[i][j],g[i][k]+g[k][j]);//Floyd
14         F(i,1,end)F(j,1,n)dp[i][j]=inf;
15         dp[1][1]=0;
16         F(i,1,end)F(j,1,n)if(dp[i][j]!=inf)F(k,2,n)
17         if(k==j||i&1<<k-1)continue;
18         else up(dp[i|1<<k-1][k],dp[i][j]+g[j][k]);
19         F(i,1,n)up(ans,dp[end][i]+g[i][1]);
20         printf("%d\n",ans);
21     }
22     return 0;
23 }
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posted @ 2016-07-02 11:34  bin_gege  阅读(165)  评论(0编辑  收藏  举报