实现详见：https://www.geeksforgeeks.org/dijkstras-shortest-path-algorithm-using-priority_queue-stl/
需要注意的是，priority_queue并无法更新内部的元素，因此我们更新dist的同时，直接把新的距离加入pq即可。pq里虽然有outdated的dist，但是由于距离过长，他们并不会更新dist。
// If there is shorted path to v through u. if (dist[v] > dist[u] + weight) { // Updating distance of v dist[v] = dist[u] + weight; pq.push(make_pair(dist[v], v)); }
时间复杂度 O(ElogV)
本题的实质是 Dijkstra’s Shortest Path Algorithm，只不过追加了一个约束条件step。
class Solution { public: typedef tuple<int,int,int> ti; // (dist,u,step) struct edge{ int end; int weight; }; int findCheapestPrice(int n, vector<vector<int>>& flights, int src, int dst, int K) { vector<vector<edge>> graph(n,vector<edge>()); for (auto flight:flights){ graph[flight[0]].push_back({flight[1],flight[2]}); } priority_queue<ti,vector<ti>,greater<ti>> q; q.emplace(0,src,K+1); while (!q.empty()){ auto [dist,u,step]=q.top(); q.pop(); if (u==dst) return dist; if (step==0) continue; for (auto [v,w]:graph[u]){ q.emplace(dist+w,v,step-1); } } return -1; } };
Dijkstra’s Shortest Path Algorithm / LeetCode 787. Cheapest Flights Within K Stops
原文：https://www.cnblogs.com/hankunyan/p/11518287.html