(简单) POJ 1847 Tram，Dijkstra。

　　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.

　　题目就是求最短路，然后 建边 i 和 第一个连边，边权为0，其他的边权为1.

代码如下：

#include <stdio.h>
#include <string.h>
#include <iostream>
#include <algorithm>
#include <vector>
#include <queue>
#include <set>
#include <map>
#include <string>
#include <math.h>
#include <stdlib.h>
#include <time.h>

using namespace std;

const int MaxN=110;
const int INF=10e8;

int vis[MaxN];

void Dijkstra(int lowcost[],int cost[][MaxN],int N,int start)
{
    int minn,minp;

    for(int i=1;i<=N;++i)
    {
        lowcost[i]=INF;
        vis[i]=0;
    }
    lowcost[start]=0;

    for(int cas=1;cas<=N;++cas)
    {
        minn=INF;
        minp=-1;
        for(int i=1;i<=N;++i)
            if(!vis[i] && lowcost[i]<minn)
            {
                minn=lowcost[i];
                minp=i;
            }

        if(minp==-1)
            return;
        vis[minp]=1;

        for(int i=1;i<=N;++i)
            if(!vis[i] && cost[minp][i]!=-1 && lowcost[i]>lowcost[minp]+cost[minp][i])
                lowcost[i]=lowcost[minp]+cost[minp][i];
    }
}

int map1[MaxN][MaxN];
int ans[MaxN];
int N,A,B;

int main()
{
    //freopen("in.txt","r",stdin);
    //freopen("out.txt","w",stdout);

    int k,a;

    for(int i=1;i<=MaxN;++i)
        for(int j=1;j<=MaxN;++j)
            map1[i][j]=-1;

    scanf("%d %d %d",&N,&A,&B);

    for(int i=1;i<=N;++i)
    {
        scanf("%d",&k);

        if(k==0)
            continue;

        scanf("%d",&a);

        map1[i][a]=0;
        --k;

        while(k--)
        {
            scanf("%d",&a);

            if(map1[i][a]==-1)
                map1[i][a]=1;
        }
    }

    //for(int i=1;i<=N;++i)
    //    for(int j=1;j<=N;++j)
    //        cout<<map1[i][j]<<‘ ‘;

    Dijkstra(ans,map1,N,A);

    printf("%d\n",ans[B]==INF ? -1 : ans[B]);

    return 0;
}