Tram POJ - 1847

题目链接：https://vjudge.net/problem/POJ-1847

思路：想从A到B使用开关少，想清楚了就是个简单的最短路，可以把不用开开关为权值0，

要开开关为权值1，就是求A到B开开关最少的次数，题目说了，每行第一个点是第 i-th点和他正好数开关开的方向连接。

 #include <iostream>
 #include <cstdio>
 #include <cstring>
 #include <algorithm>
 #include <queue>
 #include <stack>
 #include <string>
 #include <map>
 #include <cmath>
 #include <iomanip>
 using namespace std;

 typedef long long LL;
 #define inf 1e9
 #define rep(i,j,k) for(int i = (j); i <= (k); i++)
 #define rep__(i,j,k) for(int i = (j); i < (k); i++)
 #define per(i,j,k) for(int i = (j); i >= (k); i--)
 #define per__(i,j,k) for(int i = (j); i > (k); i--)

 const int N = ;
 int head[N];
 bool vis[N];
 int dis[N];
 int cnt;
 int n,s,t;

 struct Edge{
     int to;
     int w;
     int next;
 }e[N * N];

 struct node{
     int pos;
     int w;

     bool friend operator< (const node& a,const node& b){
         return a.w > b.w;
     }
 };

 void add(int u,int v,int w){
     e[cnt].to = v;
     e[cnt].w = w;
     e[cnt].next = head[u];
     head[u] = cnt++;
 }

 void dijkstra(){

     rep(i,,n) vis[i] = false;
     rep(i,,n) dis[i] = inf;
     dis[s] = ;

     priority_queue<node> que;
     que.push(node{s,});

     int pos,v,w;
     while(!que.empty()){
         pos = que.top().pos;
         que.pop();

         if(vis[pos]) continue;
         vis[pos] = true;

         for(int o = head[pos]; ~o; o = e[o].next){
             v = e[o].to;
             w = e[o].w;

             if(!vis[v] && dis[pos] + w < dis[v]){
                 dis[v] = dis[pos] + w;

                 que.push(node{v,dis[v]});
             }
         }
     }

     if(dis[t] == inf) cout << "-1" << endl;
     else cout << dis[t] << endl;
 }

 int main(){

     scanf("%d%d%d",&n,&s,&t);

     rep(i,,n) head[i] = -;
     int cnt = ;

     int tot,v;
     rep(u,,n){
         cin >> tot;

         rep(i,,tot){
             cin >> v;

             if(i == ) add(u,v,);
             else add(u,v,);
         }
     }
     dijkstra();

     getchar(); getchar();
     return ;
 }

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