图的最短路生成树

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我是靠谱客的博主疯狂蜡烛，最近开发中收集的这篇文章主要介绍图的最短路生成树，觉得挺不错的，现在分享给大家，希望可以做个参考。

概述

E. Paths and Trees

time limit per test

3 seconds

memory limit per test

256 megabytes

input

standard input

output

standard output

Little girl Susie accidentally found her elder brother's notebook. She has many things to do, more important than solving problems, but she found this problem too interesting, so she wanted to know its solution and decided to ask you about it. So, the problem statement is as follows.

Let's assume that we are given a connected weighted undirected graph G = (V, E) (here V is the set of vertices, E is the set of edges). The shortest-path tree from vertex u is such graph G1 = (V, E1) that is a tree with the set of edges E1 that is the subset of the set of edges of the initial graph E, and the lengths of the shortest paths from u to any vertex to G and to G1 are the same.

You are given a connected weighted undirected graph G and vertex u. Your task is to find the shortest-path tree of the given graph from vertex u, the total weight of whose edges is minimum possible.

Input

The first line contains two numbers, n and m (1 ≤ n ≤ 3·105, 0 ≤ m ≤ 3·105) — the number of vertices and edges of the graph, respectively.

Next m lines contain three integers each, representing an edge — ui, vi, wi — the numbers of vertices connected by an edge and the weight of the edge (ui ≠ vi, 1 ≤ wi ≤ 109). It is guaranteed that graph is connected and that there is no more than one edge between any pair of vertices.

The last line of the input contains integer u (1 ≤ u ≤ n) — the number of the start vertex.

Output

In the first line print the minimum total weight of the edges of the tree.

In the next line print the indices of the edges that are included in the tree, separated by spaces. The edges are numbered starting from 1 in the order they follow in the input. You may print the numbers of the edges in any order.

If there are multiple answers, print any of them.

Examples

input

Copy

output

Copy

2
1 2

input

Copy

output

Copy

4
2 3 4

Note

In the first sample there are two possible shortest path trees:

with edges 1 – 3 and 2 – 3 (the total weight is 3);
with edges 1 – 2 and 2 – 3 (the total weight is 2);

And, for example, a tree with edges 1 – 2 and 1 – 3 won't be a shortest path tree for vertex 3, because the distance from vertex 3 to vertex 2 in this tree equals 3, and in the original graph it is 1.

【题意】

给你一个图，然后给你一个点，求一个图，包含这个点到其他点的最短路的边的图。

要求这个图的边权和最小

【方法】

跑dijkstra，如果dis相同，选择小的边，，和普通的最短路相比，多了一个变量记录这条边的序号，，注意边*2，，

#include <bits/stdc++.h>
typedef long long ll;
using namespace std;
int Num;
char CH[20];
const int inf=0x3f3f3f3f;
struct node
{
int num,v,w,next;
}edge[600010];
struct node2
{
int u;
ll d;
bool operator <(const node2 a)const
{
return a.d<d;
}
};
priority_queue<node2> qu;
node2 A;
int T,t,n,m,Head[300010],tot;
ll dis[300010],cost[300010],use[300010],INF=1e16;
bool vis[300010],vis2[300010];
void add(int u,int v,int w,int num)
{
edge[tot].num=num;
edge[tot].v=v;
edge[tot].w=w;
edge[tot].next=Head[u];
Head[u]=tot++;
}
int main()
{
int i,j,k,u,v,w,num;
ios::sync_with_stdio(false);
ll ans;
cin>>n>>m;
memset(Head,-1,sizeof(Head));
for(i=1;i<=m;i++)
{
cin>>u>>v>>w;
add(u,v,w,i);
add(v,u,w,i);
}
for(i=1;i<=n;i++)
dis[i]=cost[i]=INF;
cin>>u;
dis[u]=0;
cost[u]=0;
A.u=u;
A.d=0;
qu.push(A);
while(!qu.empty())
{
A=qu.top();
qu.pop();
u=A.u;
if(vis[u])
continue;
vis[u]=1;
vis2[use[u]]=1;
for(i=Head[u];i!=-1;i=edge[i].next)
{
v=edge[i].v;
w=edge[i].w;
num=edge[i].num;
if(dis[u]+w<dis[v])
{
dis[v]=dis[u]+w;
cost[v]=w;
use[v]=num;
A.u=v;A.d=dis[v];
qu.push(A);
}
else if(dis[u]+w==dis[v] && w<cost[v])
{
cost[v]=w;
use[v]=num;
}
}
}
ans=0;
for(i=1;i<=m;i++)
if(vis2[i])
ans+=edge[i*2-1].w;
cout<<ans<<endl;
if(ans>0)
{
for(i=1;i<=m;i++)
if(vis2[i])
cout<<i<<" ";
cout<<endl;
}
return 0;
}