POJ 1847 Tram【Dijkstra】

Tram

Time Limit: 1000MS		Memory Limit: 30000K
Total Submissions: 13082		Accepted: 4794

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

Source

Croatia OI 2002 Regional - Juniors

 

题目大意:第一行输入一共有多少个节点，之后两个数代表起点和终点，然后接下来N行，每行第一个数k表示和当前这个点有几个点相连 ，然后随后k个数，第一个数和当前节点距离为0，其余距离为1，求起点到终点的最短距离。

思路：因为点比较少，所以所有求最短路的方法都可以通过，这里我们使用DIJ+邻接表实现。

AC代码：

#include<stdio.h>
#include<string.h>
using namespace std;
int head[100000];
struct EdgeNode
{
    int to;
    int w;
    int next;
}e[100000];
int vis[100000];
int dis[100000];
int cont,n,s,t;
void add(int from,int to,int w)
{
    e[cont].to=to;
    e[cont].w=w;
    e[cont].next=head[from];
    head[from]=cont++;
}
void Dij()
{
    memset(vis,0,sizeof(vis));
    for(int i=1;i<=n;i++)dis[i]=0x3f3f3f3f;
    dis[s]=0;
    for(int i=1;i<n;i++)
    {
        int v=-1;
        int tmp=0x3f3f3f3f;
        for(int j=1;j<=n;j++)
        {
            if(vis[j]==0&&tmp>dis[j])
            {
                tmp=dis[j];
                v=j;
            }
        }
        vis[v]=1;
        for(int k=head[v];k!=-1;k=e[k].next)
        {
            int to=e[k].to;
            int w=e[k].w;
            if(vis[to]==0&&dis[to]>dis[v]+w)
            {
                dis[to]=dis[v]+w;
            }
        }
    }
}
int main()
{
    while(~scanf("%d%d%d",&n,&s,&t))
    {
        memset(head,-1,sizeof(head));
        cont=0;
        for(int i=1;i<=n;i++)
        {
            int k;
            scanf("%d",&k);
            for(int j=0;j<k;j++)
            {
                int to;
                scanf("%d",&to);
                if(j==0)
                add(i,to,0);
                else add(i,to,1);
            }
        }
        Dij();
        if(dis[t]>=0x3f3f3f3f)
        {
            printf("-1\n");
        }
        else
        printf("%d\n",dis[t]);
    }
}