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。

  1 #include <iostream>
  2 #include <cstdio>
  3 #include <cstring>
  4 #include <algorithm>
  5 #include <queue>
  6 #include <stack>
  7 #include <string>
  8 #include <map>
  9 #include <cmath>
 10 #include <iomanip>
 11 using namespace std;
 12  
 13 typedef long long LL;
 14 #define inf 1e9
 15 #define rep(i,j,k) for(int i = (j); i <= (k); i++)
 16 #define rep__(i,j,k) for(int i = (j); i < (k); i++)
 17 #define per(i,j,k) for(int i = (j); i >= (k); i--)
 18 #define per__(i,j,k) for(int i = (j); i > (k); i--)
 19 
 20 const int N = 110;
 21 int head[N];
 22 bool vis[N];
 23 int dis[N];
 24 int cnt;
 25 int n,s,t;
 26 
 27 struct Edge{
 28     int to;
 29     int w;
 30     int next;
 31 }e[N * N];
 32 
 33 struct node{
 34     int pos;
 35     int w;
 36 
 37     bool friend operator< (const node& a,const node& b){
 38         return a.w > b.w;
 39     }
 40 };
 41 
 42 void add(int u,int v,int w){
 43     e[cnt].to = v;
 44     e[cnt].w = w;
 45     e[cnt].next = head[u];
 46     head[u] = cnt++;
 47 }
 48 
 49 void dijkstra(){
 50 
 51     rep(i,1,n) vis[i] = false;
 52     rep(i,1,n) dis[i] = inf;
 53     dis[s] = 0;
 54 
 55     priority_queue<node> que;
 56     que.push(node{s,0});
 57 
 58     int pos,v,w;
 59     while(!que.empty()){
 60         pos = que.top().pos;
 61         que.pop();
 62 
 63         if(vis[pos]) continue;
 64         vis[pos] = true;
 65 
 66         for(int o = head[pos]; ~o; o = e[o].next){
 67             v = e[o].to;
 68             w = e[o].w;
 69 
 70             if(!vis[v] && dis[pos] + w < dis[v]){
 71                 dis[v] = dis[pos] + w;
 72 
 73                 que.push(node{v,dis[v]});
 74             }
 75         }
 76     }
 77 
 78     if(dis[t] == inf) cout << "-1" << endl;
 79     else cout << dis[t] << endl;
 80 }
 81 
 82 int main(){
 83 
 84     scanf("%d%d%d",&n,&s,&t);
 85 
 86     rep(i,1,n) head[i] = -1;
 87     int cnt = 0;
 88 
 89     int tot,v;
 90     rep(u,1,n){
 91         cin >> tot;
 92 
 93         rep(i,1,tot){
 94             cin >> v;
 95 
 96             if(i == 1) add(u,v,0);
 97             else add(u,v,1);
 98         }
 99     }
100     dijkstra();
101 
102     getchar(); getchar();
103     return 0;
104 }

 

 

posted @ 2019-08-19 22:50  SummerMingQAQ  阅读(243)  评论(0)  编辑  收藏  举报