Problem A Codeforces 20C 最短路(dj,spfa)

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 

最短路
我用的DJ
之前没写出来的原因是不会保存路径

#include<stdio.h>
#include<queue>
#define LL long long
using namespace std;
const int maxn = 400005;//这个给大一点，四倍
const LL maxd = 1E13;
int v[maxn],w[maxn],next[maxn],pre[maxn],res[maxn];
int first[maxn],inq[maxn],e;
LL d[maxn];

void init()
{
    for(int i=0;i<maxn;i++)
    {
        first[i]=-1;
    }
    e=0;
}


void addeage(int x,int y,int z)
{
    v[e]=y;
    w[e]=z;
    next[e]=first[x];
    first[x]=e;
    e++;
}

void spfa(int s)
{
    queue<int> q;
    for(int i =0;i<maxn;i++)
        d[i]=maxd;
    d[s]=0;inq[s]=1;q.push(s);
    while(q.empty()==false)
    {
        int u = q.front();
        q.pop();
        inq[u]=0;
        for(int i =first[u];i!=-1;i=next[i])
        {
            if(d[v[i]]>d[u]+w[i])
            {
                d[v[i]]=(d[u]+w[i]);
                pre[v[i]]=u;//把前一个点存储
                if(inq[v[i]]==0)
                {
                    q.push(v[i]);
                    inq[v[i]]=1;
                }
            }
        }
    }
}

int main()
{
    init();
    int m,n;
    scanf("%d%d",&m,&n);
    int x,y,z;
    while(n--)
    {
        scanf("%d%d%d",&x,&y,&z);
        addeage(x,y,z);
        addeage(y,x,z);//Undirected无向图，双向的
    }
    spfa(1);
    if(d[m]==maxd)
        printf("-1");
    else
    {
        int now=m;//从终点开始把之前的一个个找出来
        int cnt=0;
        while(now!=1)
        {
           res[cnt++] = now;//用来保存路径
           now = pre[now];
        }
        res[cnt++] = 1;
        for(int i = cnt-1;i >= 0;i--)
            printf("%d ",res[i]);
    }
    return 0;
}