poj-1847-Tram

Tram

Time Limit: 1000MS		Memory Limit: 30000K
Total Submissions:17828		Accepted: 6654

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

题目大意：有N个站点，站点之间有轨道相连，但是站点同时只连向一个站点，要到该站点可以到的其它站点需要使用转换器，问从A到B需要最少使用多少次转换器

解题思路：可以将使用转换器的次数看做两站点的距离，初始化图的时候，该站点直连的站点初始化为0，其它站点初始化为1，然后由于数据量太小，随便一个最短路算法即可

代码：

#include<queue>
#include<cmath>
#include<cstdio>
#include<cstring>
#include<cstdlib>
#include<iostream>
#include<algorithm>
using namespace std;
#define ll long long
#define inf 99999999
int e[111][111];
int main()
{
    int n,a,b;
    while(scanf("%d%d%d",&n,&a,&b)!=EOF)
    {
        int k,t1,t2,t3,i,j;
        for(i=1; i<=n; i++)
        {
            for(j=1; j<=n; j++)
            {
                if(i!=j)
                    e[i][j]=inf;
                else
                    e[i][j]=0;
            }
        }
        for(i=1; i<=n; i++)
        {
            scanf("%d",&t1);
            for(j=0; j<t1; j++)
            {
                scanf("%d",&t2);
                if(j==0)//直接相连肯定为0
                    e[i][t2]=0;
                else//不相连初始化为1
                    e[i][t2]=1;
            }
        }
        for(k=1; k<=n; k++)
        {
            for(i=1; i<=n; i++)
            {
                for(j=1; j<=n; j++)
                {
                    if(e[i][j]>e[i][k]+e[k][j])
                        e[i][j]=e[i][k]+e[k][j];
                }
            }
        }
//    for(i=1; i<=n; i++)
//    {
//        for(j=1; j<=n; j++)
//            printf("%d  ",e[i][j]);
//        printf("\n");
//    }
        if(e[a][b]==inf)
            printf("-1\n");
        else
            printf("%d\n",e[a][b]);
    }
    return 0;
}