<p>Given a weighted undirected connected graph with <code>n</code>&nbsp;vertices numbered from <code>0</code> to <code>n - 1</code>,&nbsp;and an array <code>edges</code>&nbsp;where <code>edges[i] = [a<sub>i</sub>, b<sub>i</sub>, weight<sub>i</sub>]</code> represents a bidirectional and weighted edge between nodes&nbsp;<code>a<sub>i</sub></code>&nbsp;and <code>b<sub>i</sub></code>. A minimum spanning tree (MST) is a subset of the graph&#39;s edges that connects all vertices without cycles&nbsp;and with the minimum possible total edge weight.</p>

<p>Find <em>all the critical and pseudo-critical edges in the given graph&#39;s minimum spanning tree (MST)</em>. An MST edge whose deletion from the graph would cause the MST weight to increase is called a&nbsp;<em>critical edge</em>. On&nbsp;the other hand, a pseudo-critical edge is that which can appear in some MSTs but not all.</p>

<p>Note that you can return the indices of the edges in any order.</p>

<p>&nbsp;</p>
<p><strong class="example">Example 1:</strong></p>

<p><img alt="" src="https://assets.leetcode.com/uploads/2020/06/04/ex1.png" style="width: 259px; height: 262px;" /></p>

<pre>
<strong>Input:</strong> n = 5, edges = [[0,1,1],[1,2,1],[2,3,2],[0,3,2],[0,4,3],[3,4,3],[1,4,6]]
<strong>Output:</strong> [[0,1],[2,3,4,5]]
<strong>Explanation:</strong> The figure above describes the graph.
The following figure shows all the possible MSTs:
<img alt="" src="https://assets.leetcode.com/uploads/2020/06/04/msts.png" style="width: 540px; height: 553px;" />
Notice that the two edges 0 and 1 appear in all MSTs, therefore they are critical edges, so we return them in the first list of the output.
The edges 2, 3, 4, and 5 are only part of some MSTs, therefore they are considered pseudo-critical edges. We add them to the second list of the output.
</pre>

<p><strong class="example">Example 2:</strong></p>

<p><img alt="" src="https://assets.leetcode.com/uploads/2020/06/04/ex2.png" style="width: 247px; height: 253px;" /></p>

<pre>
<strong>Input:</strong> n = 4, edges = [[0,1,1],[1,2,1],[2,3,1],[0,3,1]]
<strong>Output:</strong> [[],[0,1,2,3]]
<strong>Explanation:</strong> We can observe that since all 4 edges have equal weight, choosing any 3 edges from the given 4 will yield an MST. Therefore all 4 edges are pseudo-critical.
</pre>

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

<ul>
	<li><code>2 &lt;= n &lt;= 100</code></li>
	<li><code>1 &lt;= edges.length &lt;= min(200, n * (n - 1) / 2)</code></li>
	<li><code>edges[i].length == 3</code></li>
	<li><code>0 &lt;= a<sub>i</sub> &lt; b<sub>i</sub> &lt; n</code></li>
	<li><code>1 &lt;= weight<sub>i</sub>&nbsp;&lt;= 1000</code></li>
	<li>All pairs <code>(a<sub>i</sub>, b<sub>i</sub>)</code> are <strong>distinct</strong>.</li>
</ul>

from typing import List


class UnionFind:
    # 并查集的基础模版
    def __init__(self, n):
        self.parent = [i for i in range(n)]
        self.count = n

    def find(self, x):
        while x != self.parent[x]:
            self.parent[x] = self.parent[self.parent[x]]
            x = self.parent[x]
        return x

    def union(self, x, y):
        root_x = self.find(x)
        root_y = self.find(y)
        if root_x == root_y:
            return

        self.parent[root_x] = root_y
        self.count -= 1

    def is_connected(self, x, y):
        return self.find(x) == self.find(y)


class Solution:

    def kruskal(self, edges, uf, init_mst):
        # 记录最小生成树的权重之和
        mst = init_mst

        for edge in edges:
            u, v, weight, _ = edge
            # 若这条边会产生环，则不能加入 mst
            if uf.is_connected(u, v):
                continue

            # 若这条边不会产生环，则属于最小生成树
            mst += weight
            uf.union(u, v)

        return mst, uf.count

    def findCriticalAndPseudoCriticalEdges(self, n: int, edges: List[List[int]]) -> List[List[int]]:
        edges_copy = []
        for i, edge in enumerate(edges):
            edge_copy = edge.copy()
            edge_copy.append(i)
            edges_copy.append(edge_copy)

        # 对所有边按照权重从小到大排序
        edges_copy.sort(key=lambda x: x[2])

        uf = UnionFind(n)
        mst_weight, _ = self.kruskal(edges_copy, uf, 0)

        critical = []
        pseudo_critical = []
        for i in range(len(edges_copy)):
            new_edges = edges_copy[:i] + edges_copy[i + 1:]

            uf_ignore = UnionFind(n)
            mst_ignore_weight, new_count = self.kruskal(new_edges, uf_ignore, 0)
            if new_count > 1 or mst_ignore_weight > mst_weight:
                critical.append(edges_copy[i][3])
                continue

            uf_force = UnionFind(n)
            init_mst = edges_copy[i][2]
            uf_force.union(edges_copy[i][0], edges_copy[i][1])
            mst_force_weight, _ = self.kruskal(new_edges, uf_force, init_mst)
            if mst_force_weight == mst_weight:
                pseudo_critical.append(edges_copy[i][3])

        return [critical, pseudo_critical]

883