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[101001];
int visited[101001];
vector<int> V[101001];

int bfs(){
    queue<int> Q;
    for(int i=1; i<=N+M; i++){
        dist[i]=inf;
        visited[i]=0;
    }
    dist[1]=1;
    visited[1]=1;
    Q.push(1);
    while( !Q.empty() ){
        int u=Q.front();
        if(u==N)return dist[u];
        Q.pop();
        for(int i=0; i<V[u].size(); i++){
            int v=V[u][i];
            if(!visited[v]){
                if(v<=N)
                    dist[v]=dist[u]+1;
                else
                    dist[v]=dist[u];
                Q.push(v);
                visited[v]=1;                
            }
        }
    }
    return -1;
}

int main()
{
    while( scanf("%d %d %d" ,&N ,&K ,&M)!=EOF ){
        for(int i=1; i<=M; i++){
            for(int j=0; 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
",ans);
    }
    return 0;
}