TOJ 4523 Transportation

Description

Given N stations, you want to carry goods from station 1 to station N. Among these stations, we use M tubes to connect some of them. Each tube can directly connect K stations with each other. What is the minimum number of stations to pass through to carry goods from station 1 to station N?

Input

The first line has three positive integers: N (1 ≤ N ≤ 100 000, K (1 ≤ K ≤ 1 000) and M (1 ≤ M ≤ 1 000).
Then follows M lines, each line has K positive integers describing the connected stations to this tube.

Output

Output  the minimum number of stations.

If it cannot carry goods from station 1 to station N, just output -1.

Sample Input

9 3 5
1 2 3
1 4 5
3 6 7
5 6 7
6 8 9

Sample Output

4

Source

TOJ

 

先要对图进行压缩。因为K点两两相连，则表示可以引入一个虚拟的点（N+i）这些点到的虚拟的点距离都相等。

然后进行广搜一开始想到的是SPFA这种方法。不过对于后来直接广搜就莫名奇妙的过了。

于是去问贞贞这是为什么呢？

后来想明白了，因为点点之间的距离是等距的。如果有条路到N的距离比较长的话，那么一定是晚点找到的。

所以先返回的一定是最短的。

 

 1 #include <stdio.h>
 2 #include <iostream> 
 3 #include <queue>
 4 #define inf 0x3f3f3f3f
 5 using namespace std;
 6 
 7 int N,K,M;
 8 int dist[101001];
 9 int visited[101001];
10 vector<int> V[101001];
11 
12 int bfs(){
13     queue<int> Q;
14     for(int i=1; i<=N+M; i++){
15         dist[i]=inf;
16         visited[i]=0;
17     }
18     dist[1]=1;
19     visited[1]=1;
20     Q.push(1);
21     while( !Q.empty() ){
22         int u=Q.front();
23         if(u==N)return dist[u];
24         Q.pop();
25         for(int i=0; i<V[u].size(); i++){
26             int v=V[u][i];
27             if(!visited[v]){
28                 if(v<=N)
29                     dist[v]=dist[u]+1;
30                 else
31                     dist[v]=dist[u];
32                 Q.push(v);
33                 visited[v]=1;                
34             }
35         }
36     }
37     return -1;
38 }
39 
40 int main()
41 {
42     while( scanf("%d %d %d" ,&N ,&K ,&M)!=EOF ){
43         for(int i=1; i<=M; i++){
44             for(int j=0; j<K; j++){
45                 int x;
46                 scanf("%d" ,&x);
47                 V[N+i].push_back(x);
48                 V[x].push_back(N+i);
49             }    
50         }
51         int ans=bfs();
52         printf("%d\n",ans);
53     }
54     return 0;
55 }