hdu_5418_Victor and World(状压DP+Floyd)

时间:2024-04-27 15:00:41

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。

 #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 dp[<<][],g[][],inf=<<,n,m,t,u,v,c,end,ans; int main(){
scanf("%d",&t);
while(t--){
scanf("%d%d",&n,&m),end=(<<n)-,ans=inf;
F(i,,n)F(j,,n)g[i][j]=(i==j?:inf);
F(i,,m)scanf("%d%d%d",&u,&v,&c),up(g[u][v],c),up(g[v][u],c);
F(k,,n)F(i,,n)F(j,,n)up(g[i][j],g[i][k]+g[k][j]);//Floyd
F(i,,end)F(j,,n)dp[i][j]=inf;
dp[][]=;
F(i,,end)F(j,,n)if(dp[i][j]!=inf)F(k,,n)
if(k==j||i&<<k-)continue;
else up(dp[i|<<k-][k],dp[i][j]+g[j][k]);
F(i,,n)up(ans,dp[end][i]+g[i][]);
printf("%d\n",ans);
}
return ;
}