<p>There are <code>n</code> cities connected by some number of flights. You are given an array <code>flights</code> where <code>flights[i] = [from<sub>i</sub>, to<sub>i</sub>, price<sub>i</sub>]</code> indicates that there is a flight from city <code>from<sub>i</sub></code> to city <code>to<sub>i</sub></code> with cost <code>price<sub>i</sub></code>.</p>

<p>You are also given three integers <code>src</code>, <code>dst</code>, and <code>k</code>, return <em><strong>the cheapest price</strong> from </em><code>src</code><em> to </em><code>dst</code><em> with at most </em><code>k</code><em> stops. </em>If there is no such route, return<em> </em><code>-1</code>.</p>

<p>&nbsp;</p>
<p><strong class="example">Example 1:</strong></p>
<img alt="" src="https://assets.leetcode.com/uploads/2022/03/18/cheapest-flights-within-k-stops-3drawio.png" style="width: 332px; height: 392px;" />
<pre>
<strong>Input:</strong> n = 4, flights = [[0,1,100],[1,2,100],[2,0,100],[1,3,600],[2,3,200]], src = 0, dst = 3, k = 1
<strong>Output:</strong> 700
<strong>Explanation:</strong>
The graph is shown above.
The optimal path with at most 1 stop from city 0 to 3 is marked in red and has cost 100 + 600 = 700.
Note that the path through cities [0,1,2,3] is cheaper but is invalid because it uses 2 stops.
</pre>

<p><strong class="example">Example 2:</strong></p>
<img alt="" src="https://assets.leetcode.com/uploads/2022/03/18/cheapest-flights-within-k-stops-1drawio.png" style="width: 332px; height: 242px;" />
<pre>
<strong>Input:</strong> n = 3, flights = [[0,1,100],[1,2,100],[0,2,500]], src = 0, dst = 2, k = 1
<strong>Output:</strong> 200
<strong>Explanation:</strong>
The graph is shown above.
The optimal path with at most 1 stop from city 0 to 2 is marked in red and has cost 100 + 100 = 200.
</pre>

<p><strong class="example">Example 3:</strong></p>
<img alt="" src="https://assets.leetcode.com/uploads/2022/03/18/cheapest-flights-within-k-stops-2drawio.png" style="width: 332px; height: 242px;" />
<pre>
<strong>Input:</strong> n = 3, flights = [[0,1,100],[1,2,100],[0,2,500]], src = 0, dst = 2, k = 0
<strong>Output:</strong> 500
<strong>Explanation:</strong>
The graph is shown above.
The optimal path with no stops from city 0 to 2 is marked in red and has cost 500.
</pre>

<p>&nbsp;</p>
<p><strong>Constraints:</strong></p>

<ul>
	<li><code>1 &lt;= n &lt;= 100</code></li>
	<li><code>0 &lt;= flights.length &lt;= (n * (n - 1) / 2)</code></li>
	<li><code>flights[i].length == 3</code></li>
	<li><code>0 &lt;= from<sub>i</sub>, to<sub>i</sub> &lt; n</code></li>
	<li><code>from<sub>i</sub> != to<sub>i</sub></code></li>
	<li><code>1 &lt;= price<sub>i</sub> &lt;= 10<sup>4</sup></code></li>
	<li>There will not be any multiple flights between two cities.</li>
	<li><code>0 &lt;= src, dst, k &lt; n</code></li>
	<li><code>src != dst</code></li>
</ul>

class Solution {
public:
    // bfs + graph + early termination.
    // Use vector<vector<pair<int, int>>> graph to create 
    // graph from flights. eg. graph[i] = [{j1, c1}, {j2, c2}]
    // represents start from i to j1, the cost is c1; start
    // from i to j2, the cost is c2. Use queue<pair<int, int>> q
    // for bfs, q stores {current stop, current cheapest cost from
    // src to current stop}. Push {src, 0} to the queue. 
    // Use vector<int> cost to record current cheapest cost to each 
    // stop.

    // While q is not empty, and k >= 0 (at most k stops), do level
    // traverse, namely get current queue size, then for loop the 
    // first qsize elements in the queue, these are of the same 
    // distance to reach from src to current stop. For each top
    // element in the queue, pop it, and traverse all the neighbours
    // from the graph. For each neighbour stop, calculate current cost 
    // to it, and compare it with cost[neighbour stop]. Only update
    // cost[neighbour stop] and insert {neighbour stop, cost} into 
    // the queu if current cost to neighbour stop is cheaper than 
    // existing cost[neighbour stop]. After each for loop, decrease k
    // by 1, cause we have moved 1 stop.
    // In the end, return cost[dis] if it's not INT_MAX, otherwise 
    // return -1.
    int findCheapestPrice(int n, vector<vector<int>>& flights, int src, int dst, int k) {
        vector<vector<pair<int, int>>> graph(n);
        vector<int> cost(n, INT_MAX);
        queue<pair<int, int>> q;
        q.push({src, 0});
        for (auto &f : flights) {
            graph[f[0]].push_back({f[1], f[2]});
        }
        while (!q.empty() && k >= 0) {
            int size = q.size();
            for (int i = 0; i < size; i++) {
                auto top = q.front();
                q.pop();
                for (auto &g : graph[top.first]) {
                    int cur = top.second + g.second;
                    if (cost[g.first] > cur) {
                        cost[g.first] = cur;
                        q.push({g.first, cur});
                    }
                }
            }
            k--;
        }
        return cost[dst] != INT_MAX ? cost[dst] : -1;
    }
};
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