POJ 1847(dijkstra坑爹题意）

Tram

Time Limit: 1000MS		Memory Limit: 30000K
Total Submissions: 21783		Accepted: 8107

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

题意是输入n,a,b三个数，表示有n个点(1-n)，起点是a，终点是b，然后接下来有n行，每一行的第一个数m表示后面将会有m个数，输入结构是这样的，然后我再具体的解释一下。

    3 2 1     3表示共有n个点，接下来有n行，2表示起点，1表示终点
    2 2 3     第一个数2表示后面有2个数，因为这是第1行，所以后面两个数表示从1到2和从1到3的边
    2 3 1     表示从2到3和从2到1的边
    2 1 2     表示从3到1和从3到2的边

       因为每个点都有开关，第一个点是不需要开开关的，但是之后的每一个都要开开关，意思是输入m后的第一条边是不需要开开关的，后面的2-m的边都是需要开开关的，比如样例的第二行，2 2 3就是1-2这条边不需要开开关，1-3这条边需要开开关。这道题最后问的就是从a点到b点，最少需要开多少个开关，题意很难读懂，懂了以后这道题就很简单了，把用不用开开关的作为边的权值，不用开开关的权值为0，用开开关的边权值为1，然后跑个dij就好了。

#include <iostream>
#include <cmath>
#include <cstdio>
#include <cstring>
#include <string>
#include <map>
#include <iomanip>
#include <algorithm>
#include <queue>
#include <stack>
#include <set>
#include <vector>
//const int maxn = 1e5+5;
#define ll long long
ll gcd(ll a,ll b){return b?gcd(b,a%b):a;}
ll lcm(ll a,ll b){return a/gcd(a,b)*b;}
#define inf 0x3f3f3f3f
#define MAX INT_MAX
#define FOR(i,a,b) for( int i = a;i <= b;++i)
#define bug cout<<"--------------"<<endl
using namespace std;
int n,m,tot,A,B;
int ver[11000],edge[11000],next[11000],head[11000],vis[11000],d[11000];

void add(int x,int y,int z)
{
    ver[++tot] = y,edge[tot] = z,next[tot] = head[x],head[x] = tot;
}
void dijiestra()
{
    priority_queue<pair<int,int> >que;
    memset(vis,0,sizeof(vis));
    memset(d,inf,sizeof(d));
    d[A] = 0;
    que.push(make_pair(0,A));
    while(que.size())
    {
        int x = que.top().second;que.pop();
        if(vis[x] == 1) continue;
        vis[x] = 1;
        for(int i= head[x];i;i=next[i])
        {
            int y = ver[i],z = edge[i];
            if( d[y] > d[x] + z )
            {
                d[y] = d[x] + z;
                que.push(make_pair(-d[y],y));
            }
        }
    }

}
int main()
{

      ios::sync_with_stdio(false);
    cin>>n>>A>>B;
    FOR(i,1,n)
    {
        int m,one;
        cin>>m>>one;
        add(i,one,0);
        FOR(j,2,m)
        {
            int two;
            cin>>two;
            add(i,two,1);
        }
    }
    dijiestra();
    if(d[B] == inf) cout<<"-1"<<endl;
    else cout<<d[B]<<endl;



}