Description

You are given a weighted undirected graph. The vertices are enumerated from 1 to n. Your task is to find the shortest path between the vertex 1 and the vertex n.

Input

The first line contains two integers n and m (2 ≤ n ≤ 10⁵, 0 ≤ m ≤ 10⁵), where n is the number of vertices and m is the number of edges. Following m lines contain one edge each in form a_i, b_i and w_i (1 ≤ a_i, b_i ≤ n, 1 ≤ w_i ≤ 10⁶), where a_i, b_i are edge endpoints and w_i is the length of the edge.

It is possible that the graph has loops and multiple edges between pair of vertices.

Output

Write the only integer -1 in case of no path. Write the shortest path in opposite case. If there are many solutions, print any of them.

Sample Input

Input

5 6
1 2 2
2 5 5
2 3 4
1 4 1
4 3 3
3 5 1

Output

1 4 3 5 

Input

5 6
1 2 2
2 5 5
2 3 4
1 4 1
4 3 3
3 5 1

Output

1 4 3 5 

题意：给出一张n个点m条边的无向图 (2 ≤ n ≤ 105, 0 ≤ m ≤ 105)，输出从1到n的任意一条最短路径。

解法：dijkstra或者spfa，用pre数组记录到达每个点最短距离的前驱结点。 注意：

        路径的长度范围是 1 ≤ wi ≤ 106，从1到n的最短路径长度可能超出int范围。         没有路径从1到达n时要输出-1。

#include <cstdio>
#include <cstring>
#include <queue>
#include <algorithm>
using namespace std;
typedef long long LL;
typedef pair<LL,int> pii;
const int maxn = 100010;
const int maxm = 200010;
int v[maxm],next[maxm],w[maxm];
int first[maxn],pre[maxn],res[maxn],e;
LL d[maxn];

void init(){
    e = 0;
    memset(first,-1,sizeof(first));
}

void add_edge(int a,int b,int c){
    v[e] = b;next[e] = first[a];w[e] = c;first[a] = e++;
}

void dijkstra(int src){
    priority_queue <pii,vector<pii>,greater<pii> > q;
    memset(d,-1,sizeof(d));
    d[src] = 0;
    q.push(make_pair(0,src));
    while(!q.empty()){
        while(!q.empty() && q.top().first > d[q.top().second])  q.pop();
        if(q.empty())   break;
        int u = q.top().second;
        q.pop();
        for(int i = first[u];i != -1;i = next[i]){
            if(d[v[i]] > d[u]+w[i] || d[v[i]] == -1){
                d[v[i]] = d[u]+w[i];
                q.push(make_pair(d[v[i]],v[i]));
                pre[v[i]] = u;
            }
        }
    }
}

int main(){
    init();
    int n,m,a,b,c;
    scanf("%d%d",&n,&m);
    for(int i = 0;i < m;i++){
        scanf("%d%d%d",&a,&b,&c);
        add_edge(a,b,c);
        add_edge(b,a,c);
    }
    dijkstra(1);
    int now = n,cnt = 0;
    if(d[n] == -1){
        printf("-1");
        return 0;
    }
    while(now != 1){
        res[cnt++] = now;
        now = pre[now];
    }
    res[cnt++] = 1;
    for(int i = cnt-1;i >= 0;i--)  printf("%d ",res[i]);
    return 0;
}

这题不错，正好最近在看最短路的东西，就拿它练手了~