E. Paths and Trees

Time Limit: 20 Sec Memory Limit: 256 MB

题目连接

http://codeforces.com/contest/545/problem/E

Description

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 — 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 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.

Sample Input

3 3 1 2 1 2 3 1 1 3 2 3

Sample Output

1 2

HINT

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

题解：

跑dijsktra，跑的过程中，如果dis相同，选择小的边，然后就完了……

代码：

//qscqesze#include 
     
      #include 
      
       #include 
       
        #include 
        
         #include 
         
          #include 
          
           #include 
           
            #include 
            
             #include 
             
              #include 
              
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                   typedef long long ll;using namespace std;//freopen("D.in","r",stdin);//freopen("D.out","w",stdout);#define sspeed ios_base::sync_with_stdio(0);cin.tie(0)#define maxn 200001#define mod 10007#define eps 1e-9int Num;char CH[20];//const int inf=0x7fffffff; //нчоч╢Сconst int inf=0x3f3f3f3f;/*inline void P(int x){ Num=0;if(!x){putchar('0');puts("");return;} while(x>0)CH[++Num]=x%10,x/=10; while(Num)putchar(CH[Num--]+48); puts("");}*/inline ll read(){ ll x=0,f=1;char ch=getchar(); while(ch<'0'||ch>'9'){ if(ch=='-')f=-1;ch=getchar();} while(ch>='0'&&ch<='9'){x=x*10+ch-'0';ch=getchar();} return x*f;}inline void P(int x){ Num=0;if(!x){putchar('0');puts("");return;} while(x>0)CH[++Num]=x%10,x/=10; while(Num)putchar(CH[Num--]+48); puts("");}//**************************************************************************************struct node{ int num,v,w,next;}edge[600010];struct node2{ int u; ll d; bool operator <(const node2 a)const { return a.d
                   
                     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; ll ans; scanf("%d%d",&n,&m); memset(Head,-1,sizeof(Head)); for(i=1;i<=m;i++) { scanf("%d%d%d",&u,&v,&w); add(u,v,w,i); add(v,u,w,i); } for(i=1;i<=n;i++) dis[i]=cost[i]=INF; scanf("%d",&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
                    
                     0) { for(i=1;i<=m;i++) if(vis2[i]) printf("%d ",i); printf("\n"); }}

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