<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:

    const int INF = 1e9;

    int n, m, k;
    vector<int> h, ne, e, w;
    vector<vector<int>> dist;
    vector<vector<bool>> vis;
    int idx;

    void add(int a, int b, int c) {
        ne[idx] = h[a], e[idx] = b, w[idx] = c, h[a] = idx ++;
    }

    int dijkstra(int s, int t) {
        dist = vector<vector<int>>(n, vector<int>(k + 2, INF));
        vis = vector<vector<bool>>(n, vector<bool>(k + 2, false));
        dist[s][0] = 0;
        priority_queue<vector<int>, vector<vector<int>>, greater<vector<int>>> heap;
        heap.push({0, s, 0});

        while (heap.size()) {
            auto vec = heap.top();
            heap.pop();
            int u = vec[1], d = vec[0], mv = vec[2];
            if (vis[u][mv]) continue;
            vis[u][mv] = true;
            if (mv + 1 > k + 1) continue;

            for (int i = h[u]; i != -1; i = ne[i]) {
                int j = e[i];
                if (dist[j][mv + 1] > dist[u][mv] + w[i]) {
                    dist[j][mv + 1] = dist[u][mv] + w[i];
                    heap.push({dist[j][mv + 1], j, mv + 1});
                }
            }
        }
        int res = INF;
        for (int i = 0; i <= k + 1; i ++) res = min(res, dist[t][i]);
        if (res == INF) return -1;
        return res;
    }

    int findCheapestPrice(int _n, vector<vector<int>>& flights, int src, int dst, int _k) {
        n = _n, k = _k, m = flights.size();
        h = vector<int>(n, -1); ne = e = w = vector<int>(m, 0);
        for (auto flight: flights) {
            add(flight[0], flight[1], flight[2]);
        }
        return dijkstra(src, dst);
    }
};
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