Codeforces 938D Buy a Ticket

Buy a Ticket

题意要求：求出每个城市看演出的最小费用， 注意的一点就是车票要来回的。

题解：dijkstra 生成优先队列的时候直接将在本地城市看演出的费用放入队列里， 然后直接跑就好了,  dis数组存的是， 当前情况下的最小花费是多少。

代码：

 #include<iostream>
 #include<cstring>
 #include<string>
 #include<queue>
 #include<vector>
 #include<algorithm>
 #include<cmath>
 #include<iomanip>
 #include<cstdio>
 #define LL long long
 #define ULL unsigned LL
 #define lson l,m,rt<<1
 #define rson m+1,r,rt<<1|1
 #define fi first
 #define se second
 using namespace std;
 typedef pair<LL, int> pll;
 const int N = 2e5+;
 LL dis[N];
 int head[N];
 struct Node{
     int to;
     int nt;
     LL ct;
 }e[N<<];
 priority_queue<pll, vector<pll>, greater<pll> > q;
 void dijkstra(){
     while(!q.empty()){
         int u = q.top().se;
         LL w = q.top().fi;
         q.pop();
         if(dis[u] != w) continue;
         for(int i = head[u]; ~i; i = e[i].nt){
             int v = e[i].to;
             if(dis[v] > dis[u] + e[i].ct){
                 dis[v] = dis[u] + e[i].ct;
                 q.push(pll(dis[v],v));
             }
         }
     }
 }
 int tot = ;
 void add(int u, int v, LL w){
     e[tot].ct = w;
     e[tot].to = v;
     e[tot].nt = head[u];
     head[u] = tot++;
 }
 int main(){
     ios::sync_with_stdio(); cin.tie(); cout.tie();
     memset(head, -, sizeof(head));
     int n, m;
     cin >> n >> m;
     int u, v;
     LL ct;
     for(int i = ; i <= m; i++){
         cin >> u >> v >> ct;
         add(u,v,ct*);
         add(v,u,ct*);
     }
     for(int i = ; i <= n; i++){
         cin >> ct;
         q.push(pll(ct,i));
         dis[i] = ct;
     }
     dijkstra();
     for(int i = ; i < n; i++){
         cout << dis[i] << ' ';
     }
     cout << dis[n] << endl;
     return ;
 }