Codeforces Round #303 (Div. 2)E. Paths and Trees 最短路

E. Paths and Trees

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 graphG = (V, E) (here V is the set of vertices, E is the set of edges). The shortest-path tree from vertexu is such graph G₁ = (V, E₁) that is a tree with the set of edgesE₁ that is the subset of the set of edges of the initial graphE, and the lengths of the shortest paths fromu to any vertex to G and to G₁ are the same.

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

Input

The first line contains two numbers, n andm (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 —u_i, v_i, w_i — the numbers of vertices connected by an edge and the weight of the edge (u_i ≠ v_i, 1 ≤ w_i ≤ 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 from1 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.

Sample test(s)

Input

3 31 2 12 3 11 3 23

Output

21 2 

Input

4 41 2 12 3 13 4 14 1 24

Output

42 3 4 

Note

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

• with edges 1 – 3 and 2 – 3 (the total weight is3);
• with edges 1 – 2 and 2 – 3 (the total weight is2);

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 vertex3 to vertex 2 in this tree equals3, and in the original graph it is 1.

题意：用最短路构造最小生成树，输出总的权值是多少，用到了那些边

思路：SPFA求出单源最短路，并记下用到了那些边，然后统计输出答案，值得注意的是，最短路有多条的时候

#include<bits/stdc++.h>using namespace std;typedef long long LL;const int maxn=300010;const LL INF=1e16;struct E{    int v,next,w,id;}edge[maxn*2];int N,M;int st;LL dis[maxn];bool vis[maxn];int W[maxn];int head[maxn],tot;int pre[maxn],cnt[maxn];void init(){    tot=0;    memset(head,-1,sizeof(head));    memset(pre,0,sizeof(pre));    memset(cnt,0,sizeof(cnt));}void add_edge(int u,int v,int w,int id){    edge[tot].v=v;    edge[tot].next=head[u];    edge[tot].w=w;    edge[tot].id=id;    head[u]=tot++;}void SPFA(){    for(int i=0;i<=N;i++)        dis[i]=INF,vis[i]=0;    queue<int> q;    q.push(st);    dis[st]=0;    vis[st]=1;    while(!q.empty())    {        int u=q.front();q.pop();        vis[u]=0;        for(int i=head[u];i!=-1;i=edge[i].next)        {            int v=edge[i].v;            int w=edge[i].w;            if(dis[v]>dis[u]+w)            {                dis[v]=dis[u]+w;                pre[v]=edge[i].id;                //pre[v]=make_pair(u,edge[i].id);                if(!vis[v])                {                    q.push(v);                    vis[v]=1;                }            }            else if(dis[v]==dis[u]+w)            {                if(w<W[pre[v]])                {                    pre[v]=edge[i].id;                    if(!vis[v])                    {                        q.push(v);                        vis[v]=1;                    }                }            }        }    }}int main(){    scanf("%d%d",&N,&M);    init();    for(int i=1;i<=M;i++)    {        int u,v,w;        scanf("%d%d%d",&u,&v,&w);        add_edge(u,v,w,i);        add_edge(v,u,w,i);        W[i]=w;    }    scanf("%d",&st);    SPFA();    memset(vis,0,sizeof(vis));    for(int i=1;i<=N;i++)        vis[pre[i]]=1;    LL sum=0;    for(int i=1;i<=N;i++)        if(pre[i]>0)sum+=W[pre[i]];    cout<<sum<<endl;    for(int i=1;i<=N;i++)        if(pre[i])printf("%d ",pre[i]);    printf("\n");    return 0;}