poj1847 Tram（最短路dijkstra）

描述：

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：

　　有n个点，并且给了起点和终点ab，接下来的n行的第一个数k表示与第i个点相连的点数，每一行接下来有k个数，表示与i点相连接的点，并且在这k个点中，第一个点可直接到达，及路径长度为0，其他的点需要改一次扳手，路径长度为1，用dijkstra即可解决。

代码：

#include <iostream>
#include <stdio.h>

using namespace std;
#define inf 100100100
bool vis[105];
int n,a,b;
int map[105][105];
int d[105];

void dijkstra()
{
    int i,j,v,f;
    for(i=1;i<=n;i++)
    {
        vis[i]=0;
        d[i]=map[a][i];
    }
    vis[a]=1;
    d[a]=0;
    for(i=1;i<n;i++)
    {
        f=inf;v=a;
        for(j=1;j<=n;j++)
        {
            if(d[j]<inf&&!vis[j]&&d[j]<f)
            {
                f=d[j];
                v=j;
            }
        }
        if(f>inf) break;
        vis[v]=1;
        for(j=1;j<=n;j++)
            if(!vis[j]&&map[v][j]<inf&&d[v]+map[v][j]<d[j])
                d[j]=d[v]+map[v][j];
    }
}

int main()
{
    int k,m;
    scanf("%d%d%d",&n,&a,&b);
    for(int i=1;i<=n;i++)
        for(int j=1;j<=n;j++)
        {
            if(i==j) map[i][j]=0;
            else map[i][j]=inf;
        }
    for(int i=1;i<=n;i++)
    {
        scanf("%d",&k);
        for(int j=0;j<k;j++)
        {
            scanf("%d",&m);
            if(j==0) map[i][m]=0;
            else map[i][m]=1;
        }
    }
    dijkstra();
    if(d[b]>=inf) cout<<"-1"<<endl;
    else
        cout<<d[b]<<endl;
    return 0;
}