POJ 1847：Tram

Tram

Time Limit: 1000MS		Memory Limit: 30000K
Total Submissions: 11771		Accepted: 4301

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

这题的题意是给了N个交叉口，每个交叉口有自己能转到的交叉口。注意这里：First number in the i-th line, Ki (0 <= Ki <= N-1), represents the number of rails going out of the i-th intersection.即每一行的第二个数字代表该交叉口默认的通向，是不需要手动转的，剩下的交叉口想去的话都需要手动转一次。现在想要从A口走到B口，走的路程想要转的次数时最少的，问最少转的值。

每一行的第2个数字，其距离为0。其余的距离设置为1。

代码：

#include <iostream>
#include <algorithm>
#include <cmath>
#include <vector>
#include <string>
#include <cstring>
#pragma warning(disable:4996)
using namespace std;

int num;
int dis[1005][1005];

void init()
{
	int i,j;
	for(i=1;i<=num;i++)
	{
		for(j=1;j<=num;j++)
		{
			if(i==j)
			{
				dis[i][j]=0;
			}
			else
			{
				dis[i][j]=1005;
			}
		}
	}
}
int main()
{
	int i,j,k,i_num,x,y;
	cin>>num>>x>>y;

	init();
	for(i=1;i<=num;i++)
	{
		cin>>i_num;
		int h;
		for(j=1;j<=i_num;j++)
		{
			cin>>h;
			if(j==1)//第一个数字代表原来的方向，不需要转
				dis[i][h]=0;
			else    //之后代表要转
				dis[i][h]=1;
		}
	}
	for(k=1;k<=num;k++)
	{
		for(i=1;i<=num;i++)
		{
			for(j=1;j<=num;j++)
			{
				if(dis[i][k]+dis[k][j]<dis[i][j])
				{
					dis[i][j]=dis[i][k]+dis[k][j];
				}
			}
		}
	}
	if(dis[x][y]>=1000)
		cout<<-1<<endl;
	else
		cout<<dis[x][y]<<endl;
	return 0;
}

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