poj 1847

Tram

Time Limit: 1000MS		Memory Limit: 30000K
Total Submissions: 8450		Accepted: 3059

Description

Tram network in Zagreb consists of a number of intersections and rails connecting some of them. In every intersection there is a switch pointing to the one of the rails going out of the intersection. When the tram enters the intersection it can leave only in the direction the switch is pointing. If the driver wants to go some other way, he/she has to manually change the switch. 

When a driver has do drive from intersection A to the intersection B he/she tries to choose the route that will minimize the number of times he/she will have to change the switches manually. 

Write a program that will calculate the minimal number of switch changes necessary to travel from intersection A to intersection B. 

Input

The first line of the input contains integers N, A and B, separated by a single blank character, 2 <= N <= 100, 1 <= A, B <= N, N is the number of intersections in the network, and intersections are numbered from 1 to N. 

Each of the following N lines contain a sequence of integers separated by a single blank character. First number in the i-th line, Ki (0 <= Ki <= N-1), represents the number of rails going out of the i-th intersection. Next Ki numbers represents the intersections directly connected to the i-th intersection.Switch in the i-th intersection is initially pointing in the direction of the first intersection listed. 

Output

The first and only line of the output should contain the target minimal number. If there is no route from A to B the line should contain the integer "-1".

Sample Input

3 2 1
2 2 3
2 3 1
2 1 2

Sample Output

0

这是一道dijkstra的模板题，这个题建图的关键点就在于把道路间开关的搬动次数当做相邻点的间的距离，以此来算出最短距离 即最少的次数。

下面是代码：

#include <cstdio>
#include <cstring>
#include <iostream>
#include <algorithm>

using namespace std;

const int inf=1000000000;
const int maxn=110;

int w[maxn][maxn];//w[i][j]表示从i到j是否联通，0代表不需要，1代表需要，inf代表不连通
bool vis[maxn];//标记节点是否使用     
int dist[maxn];//标记从起点到i的距离
int  n,start,end;

void Dijkstra(){
    memset(vis,false,sizeof(vis));
    for(int i=1;i<=n;i++)
        dist[i]=w[start][i];
    dist[start]=0;
    vis[start]=true;
    for(int i=1;i<=n;i++){
         int x,m=inf;
         for(int y=1;y<=n;y++){
              if(!vis[y]&&dist[y]<=m)
                   m=dist[x=y];
         }
         vis[x]=true;
         for(int y=1;y<=n;y++){
              dist[y]=min(dist[y],dist[x]+w[x][y]);
         }
     }
     if(dist[end]==inf) printf("-1\n");
     else printf("%d\n",dist[end]);
}

int main(){
     while(scanf("%d%d%d",&n,&start,&end)!=EOF){
          for(int i=1;i<=n;i++){
              dist[i]=inf;
              for(int j=1;j<=n;j++)
                  w[i][j]=inf;
          }
          int path,y;
          for(int i=1;i<=n;i++){
             scanf("%d",&path);
             for(int j=1;j<=path;j++){
                 scanf("%d",&y);
                 if(j==1) w[i][y]=0;
                 else  w[i][y]=1;
              }
          }
          Dijkstra();
    }
    return 0;
}