P1629 and p1629

P1629 and p1629
Description

There is a postman to deliver, the post office at Node 1. He wants to send N-1 in total, the destination is 2 ~ N. Because the traffic in this city is busy, all the roads are single lines. There are M roads in total. It takes some time to pass through each road. The postman can only carry one thing at a time. How long does it take to deliver this N-1 and finally return to the post office.

Input/Output Format

Input Format:

 

The first row contains two integers, N and M.

From row 2nd to row M + 1, each line has three numbers U, V, and W, indicating that there is A road from A to B that takes W time. 1 <= U, V <= N, 1 <= W <= 10000. Input ensures that any two points can reach each other.

[Data scale]

For 30% of data, 1 ≤ N ≤ 200;

For 100% of data, 1 ≤ N ≤ 100000, 1 ≤ M ≤.

 

Output Format:

 

The output contains only one row and an integer, which is the minimum time required.

 

Input and Output sample input sample #1:

5 102 3 51 5 53 5 61 2 81 3 85 3 44 1 84 5 33 5 65 4 2

Output sample #1:

83

First, find the shortest length from Point 1 to other points,
Store all edges in reverse order.
Try again

 1 #include<iostream> 2 #include<cstdio> 3 #include<cstring> 4 #include<cmath> 5 #include<algorithm> 6 #include<queue> 7 #define lli long long int  8 using namespace std; 9 const int MAXN=100001;10 const int maxn=0x7ffff;11 void read(int &n)12 {13     char c='+';int x=0;bool flag=0;14     while(c<'0'||c>'9')15     {c=getchar();if(c=='-')flag=1;}16     while(c>='0'&&c<='9')17     {x=x*10+(c-48);c=getchar();}18     flag==1?n=-x:n=x;19 }20 int n,m;21 struct node22 {23     int u,v,w,nxt;24 }edge[MAXN*4];25 int head[MAXN];26 int num=1;27 int x[MAXN],y[MAXN],z[MAXN];28 void add_edge(int x,int y,int z)29 {30     edge[num].u=x;31     edge[num].v=y;32     edge[num].w=z;33     edge[num].nxt=head[x];34     head[x]=num++;35 }36 int vis[MAXN];37 int dis[MAXN];38 void dj(int bg)39 {40     for(int i=1;i<=n;i++)41         dis[i]=maxn;42     dis[bg]=0;43     queue<int>q;44     memset(vis,0,sizeof(vis));45     q.push(1);46     for(int i=1;i<=n;i++)47     {48         int nowmin=maxn;49         int pos=-1;50         for(int i=1;i<=n;i++)51         {52             if(vis[i]==0&&dis[i]<nowmin)53             {54                 pos=i;55                 nowmin=dis[i];56             }57         }    58         vis[pos]=1;59         for(int i=head[pos];i!=-1;i=edge[i].nxt)60         {61             int will=edge[i].v;62             if(dis[will]>dis[edge[i].u]+edge[i].w)63                 dis[will]=dis[edge[i].u]+edge[i].w;64         }65     }66 }67 int main()68 {69     read(n);read(m);70     for(int i=1;i<=n;i++)71         head[i]=-1;72     for(int i=1;i<=m;i++)73     {74         75         read(x[i]);76         read(y[i]);77         read(z[i]);78         add_edge(x[i],y[i],z[i]);79     }80     int ans=0;81     dj(1);82     int ans2=0;83     for(int i=1;i<=n;i++)84         ans+=dis[i];85     for(int i=1;i<=n;i++)86         head[i]=-1;87     num=1;88     for(int i=1;i<=m;i++)89         add_edge(y[i],x[i],z[i]);90     dj(1);91     for(int i=1;i<=n;i++)92         ans+=dis[i];93     printf("%d",ans);94     return 0;95 }