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leetcode-cn/problem (Chinese)/K 站中转内最便宜的航班 [cheapest-flights-within-k-stops].html
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<p>&nbsp;</p>
<p><strong>示例 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>输入:</strong>
n = 3, edges = [[0,1,100],[1,2,100],[0,2,500]]
src = 0, dst = 2, k = 1
<strong>输出:</strong> 200
<strong>解释:</strong>
城市航班图如下
<img alt="" src="https://s3-lc-upload.s3.amazonaws.com/uploads/2018/02/16/995.png" style="height: 180px; width: 246px;" />
从城市 0 到城市 2 在 1 站中转以内的最便宜价格是 200，如图中红色所示。</pre>
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>输出:</strong> 700
<strong>解释:</strong> 城市航班图如上
从城市 0 到城市 3 经过最多 1 站的最佳路径用红色标记，费用为 100 + 600 = 700。
请注意，通过城市 [0, 1, 2, 3] 的路径更便宜，但无效，因为它经过了 2 站。
</pre>
<p><strong>示例 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>输入:</strong>
n = 3, edges = [[0,1,100],[1,2,100],[0,2,500]]
src = 0, dst = 2, k = 0
<strong>输出:</strong> 500
n = 3, edges = [[0,1,100],[1,2,100],[0,2,500]], src = 0, dst = 2, k = 1
<strong>输出:</strong> 200
<strong>解释:</strong>
城市航班图如下
<img alt="" src="https://s3-lc-upload.s3.amazonaws.com/uploads/2018/02/16/995.png" style="height: 180px; width: 246px;" />
城市航班图如上
从城市 0 到城市 2 经过最多 1 站的最佳路径标记为红色，费用为 100 + 100 = 200。
</pre>
从城市 0 到城市 2 在 0 站中转以内的最便宜价格是 500，如图中蓝色所示。</pre>
<p>&nbsp;</p>
<p><strong class="example">示例 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>
<b>输入：</b>n = 3, flights = [[0,1,100],[1,2,100],[0,2,500]], src = 0, dst = 2, k = 0
<b>输出：</b>500
<strong>解释：</strong>
城市航班图如上
从城市 0 到城市 2 不经过站点的最佳路径标记为红色，费用为 500。
</pre>
<p><strong>提示：</strong></p>