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 ai, bi and wi (1 ≤ ai, bi ≤ n, 1 ≤ wi ≤ 106), where ai, bi are edge endpoints and wi 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
 1 #include<stdio.h>
 2 #include<queue>
 3 #define LL long long
 4 using namespace std;
 5 const int maxn = 400005;//这个给大一点，四倍
 6 const LL maxd = 1E13;
 7 int v[maxn],w[maxn],next[maxn],pre[maxn],res[maxn];
 8 int first[maxn],inq[maxn],e;
 9 LL d[maxn];
10 
11 void init()
12 {
13     for(int i=0;i<maxn;i++)
14     {
15         first[i]=-1;
16     }
17     e=0;
18 }
19 
20 
21 void addeage(int x,int y,int z)
22 {
23     v[e]=y;
24     w[e]=z;
25     next[e]=first[x];
26     first[x]=e;
27     e++;
28 }
29 
30 void spfa(int s)
31 {
32     queue<int> q;
33     for(int i =0;i<maxn;i++)
34         d[i]=maxd;
35     d[s]=0;inq[s]=1;q.push(s);
36     while(q.empty()==false)
37     {
38         int u = q.front();
39         q.pop();
40         inq[u]=0;
41         for(int i =first[u];i!=-1;i=next[i])
42         {
43             if(d[v[i]]>d[u]+w[i])
44             {
45                 d[v[i]]=(d[u]+w[i]);
46                 pre[v[i]]=u;//把前一个点存储
47                 if(inq[v[i]]==0)
48                 {
49                     q.push(v[i]);
50                     inq[v[i]]=1;
51                 }
52             }
53         }
54     }
55 }
56 
57 int main()
58 {
59     init();
60     int m,n;
61     scanf("%d%d",&m,&n);
62     int x,y,z;
63     while(n--)
64     {
65         scanf("%d%d%d",&x,&y,&z);
66         addeage(x,y,z);
67         addeage(y,x,z);//Undirected无向图，双向的
68     }
69     spfa(1);
70     if(d[m]==maxd)
71         printf("-1");
72     else
73     {
74         int now=m;//从终点开始把之前的一个个找出来
75         int cnt=0;
76         while(now!=1)
77         {
78            res[cnt++] = now;//用来保存路径
79            now = pre[now];
80         }
81         res[cnt++] = 1;
82         for(int i = cnt-1;i >= 0;i--)
83             printf("%d ",res[i]);
84     }
85     return 0;
86 }
87

 



posted @ 2014-08-04 08:46  Run_dream  阅读(425)  评论(0)  编辑  收藏  举报