Happy New Year

There are n stations in Mianyang City, and m two-way roads connect some of them. Each two stations can be connected by at most one road. Any station can be connected by one or more roads to other stations, but different routes may take different time. The time spent on a path is the sum of the time required for all roads on the path. Jia's home is at station 1. He has five relatives, namely, Station A, B, C, D, and E. After the Chinese New Year, he needs to start from his own home, visit every relative (in any order), and send them holiday greetings. How does one take the least time? Input

Line 1: n (n <= 50,000), m (M <= 100,000) indicates the number of stations and highways. Row 2: A, B, C, D, E, number of the station where the five relatives are located (1 <= N ). The following m rows have three integers x, y, t (1 <= X, Y <= N, 1 <= T <= 100 ), number and time of the two stations connected by the road.

Output

Only one row contains an integer T, which is the minimum total time.

Sample Input

6 62 3 4 5 61 2 82 3 33 4 44 5 55 6 21 6 7

Sample output

21

#include<cstdio>#include<algorithm>#include<cstring>#include<new>#include<queue>#include<iostream>using namespace std;int s[7];int dis[7][50001];int n;int m,u,v,w,ans=2e9;bool pg[50001];int flag[7];int ll[7];int ne[300001],fir[50001],val[300001],to[300001],tot;void add(int u,int v,int w){    to[++tot]=v;val[tot]=w;ne[tot]=fir[u];fir[u]=tot;}queue<int> q;int doit(){    int lol=0;    for(int i=1;i<=6;i++)    {        lol+=dis[ll[i]][s[ll[i+1]]];    }    ans=min(lol,ans);}void dfs(int x){    if(x==6) doit();    for(int i=2;i<=6;i++)    {        if(!flag[i])        {            flag[i]=1;            ll[x+1]=i;            dfs(x+1);            flag[i]=0;            ll[x+1]=0;        }    }}int main(){    cin>>n>>m;    s[1]=1;    ll[1]=1;    for(int i=2;i<=6;i++) cin>>s[i];    for(int i=1;i<=m;i++)    {        cin>>u>>v>>w;        add(u,v,w);        add(v,u,w);    }    for(int i=1;i<=6;i++)    {        while(!q.empty())q.pop();        for(int j=1;j<=n;j++) dis[i][j]=1e8;        for(int j=1;j<=n;j++) pg[j]=0;        q.push(s[i]);dis[i][s[i]]=0;pg[s[i]]=1;        while(!q.empty())        {            int xx=q.front();            q.pop();pg[xx]=0;            for(int k=fir[xx];k;k=ne[k])            {                int ttt=to[k];                if(dis[i][xx]+val[k]<=dis[i][ttt])                {                    dis[i][ttt]=dis[i][xx]+val[k];                    if(!pg[ttt])                    {                        q.push(ttt);                        pg[ttt]=1;                    }                }            }        }    }    dfs(1);    cout<<ans<<endl;}

 

Happy New Year