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。

 #include <stdio.h>
 #include <iostream>
 #include <queue>
 #define inf 0x3f3f3f3f
 using namespace std;

 int N,K,M;
 int dist[];
 int visited[];
 vector<int> V[];

 int bfs(){
     queue<int> Q;
     for(int i=; i<=N+M; i++){
         dist[i]=inf;
         visited[i]=;
     }
     dist[]=;
     visited[]=;
     Q.push();
     while( !Q.empty() ){
         int u=Q.front();
         if(u==N)return dist[u];
         Q.pop();
         for(int i=; i<V[u].size(); i++){
             int v=V[u][i];
             if(!visited[v]){
                 if(v<=N)
                     dist[v]=dist[u]+;
                 else
                     dist[v]=dist[u];
                 Q.push(v);
                 visited[v]=;
             }
         }
     }
     return -;
 }

 int main()
 {
     while( scanf("%d %d %d" ,&N ,&K ,&M)!=EOF ){
         for(int i=; i<=M; i++){
             for(int j=; j<K; j++){
                 int x;
                 scanf("%d" ,&x);
                 V[N+i].push_back(x);
                 V[x].push_back(N+i);
             }
         }
         int ans=bfs();
         printf("%d\n",ans);
     }
     return ;
 }