Tram POJ

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 Outpu

0

翻译：有n个点，从a到b。从一个点到另一个点之间是铁路，铁路有初始的指向，如果要改变指向，则需要搬动扳手
，问从a到b最少需要搬动几次扳手。输入第一行为n,a,b;往下n行，每行第一个数表示此点链接的轨道数，第二个数为初始轨道指向的点
，然后是其他轨道指向的点。

思路：最短路，及那个初始指向设为0，其他指向设为1，不指向的设为无穷大，然后根据最短路算法的出从a到b之间安定最小数即可。结果需要解一个判断是否有最小的答案

代码：

#include <cstdio>
#include <fstream>
#include <algorithm>
#include <cmath>
#include <deque>
#include <vector>
#include <queue>
#include <string>
#include <cstring>
#include <map>
#include <stack>
#include <set>
#include <sstream>
#include <iostream>
#define mod 998244353
#define eps 1e-6
#define ll long long
#define INF 0x3f3f3f3f
using namespace std;

int n,a,b;
//ma表示两点之间的距离
int ma[105][105];
//dis表示起点到其他点之间的距离
int dis[105];
//vis表示当前点的dis是否为最短的
int vis[105];
 void dijkstra()
{
    //初始化标记
    memset(vis,0,sizeof(vis));
    //标记起点
    vis[a]=1;
    //初始化初距离
    for(int i=1;i<=n;i++)
    {
        dis[i]=ma[a][i];
    }
    //起点为0；
    dis[a]=0;
    for(int i=1;i<n;i++)
    {
        int mi=INF;
        int k;
        //找当前起点到各点之间的最短的
        for(int j=1;j<=n;j++)
        {
            if(!vis[j]&&dis[j]<mi)
            {
                mi=dis[j];
                k=j;
            }
        }
        //标记起点
        vis[k]=1;
        //更新所有的dis数据
        for(int j=1;j<=n;j++)
        {
            if(dis[j]>dis[k]+ma[k][j])
            {
                dis[j]=dis[k]+ma[k][j];
            }
        }

    }
}
int main()
{
    while(~scanf("%d %d %d",&n,&a,&b))
    {
        //初始化两点之间的距离
        memset(ma,INF,sizeof(ma));
        for(int i=1;i<=n;i++)
        {
            //一样的点之间距离为0
            ma[i][i]=0;
            int c,d;
            scanf("%d",&c);
            for(int j=1;j<=c;j++)
            {
                scanf("%d",&d);
                //第一个输入表示不需要变道
                if(j==1)
                {
                    ma[i][d]=0;
                }//往后都需要变道
                else
                {
                    ma[i][d]=1;
                }
            }
        }
        dijkstra();
        //如果无解，则输出-1.否则输出解
        if(dis[b]>=INF)
        {
            printf("-1
");
        }
        else
        {
            printf("%d
",dis[b]);
        }
    }
}