<p>You are given an array <code>pairs</code>, where <code>pairs[i] = [x<sub>i</sub>, y<sub>i</sub>]</code>, and:</p>

<ul>
	<li>There are no duplicates.</li>
	<li><code>x<sub>i</sub> &lt; y<sub>i</sub></code></li>
</ul>

<p>Let <code>ways</code> be the number of rooted trees that satisfy the following conditions:</p>

<ul>
	<li>The tree consists of nodes whose values appeared in <code>pairs</code>.</li>
	<li>A pair <code>[x<sub>i</sub>, y<sub>i</sub>]</code> exists in <code>pairs</code> <strong>if and only if</strong> <code>x<sub>i</sub></code> is an ancestor of <code>y<sub>i</sub></code> or <code>y<sub>i</sub></code> is an ancestor of <code>x<sub>i</sub></code>.</li>
	<li><strong>Note:</strong> the tree does not have to be a binary tree.</li>
</ul>

<p>Two ways are considered to be different if there is at least one node that has different parents in both ways.</p>

<p>Return:</p>

<ul>
	<li><code>0</code> if <code>ways == 0</code></li>
	<li><code>1</code> if <code>ways == 1</code></li>
	<li><code>2</code> if <code>ways &gt; 1</code></li>
</ul>

<p>A <strong>rooted tree</strong> is a tree that has a single root node, and all edges are oriented to be outgoing from the root.</p>

<p>An <strong>ancestor</strong> of a node is any node on the path from the root to that node (excluding the node itself). The root has no ancestors.</p>

<p>&nbsp;</p>
<p><strong class="example">Example 1:</strong></p>
<img src="https://assets.leetcode.com/uploads/2020/12/03/trees2.png" style="width: 208px; height: 221px;" />
<pre>
<strong>Input:</strong> pairs = [[1,2],[2,3]]
<strong>Output:</strong> 1
<strong>Explanation:</strong> There is exactly one valid rooted tree, which is shown in the above figure.
</pre>

<p><strong class="example">Example 2:</strong></p>
<img alt="" src="https://assets.leetcode.com/uploads/2020/12/03/tree.png" style="width: 234px; height: 241px;" />
<pre>
<strong>Input:</strong> pairs = [[1,2],[2,3],[1,3]]
<strong>Output:</strong> 2
<strong>Explanation:</strong> There are multiple valid rooted trees. Three of them are shown in the above figures.
</pre>

<p><strong class="example">Example 3:</strong></p>

<pre>
<strong>Input:</strong> pairs = [[1,2],[2,3],[2,4],[1,5]]
<strong>Output:</strong> 0
<strong>Explanation:</strong> There are no valid rooted trees.</pre>

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

<ul>
	<li><code>1 &lt;= pairs.length &lt;= 10<sup>5</sup></code></li>
	<li><code>1 &lt;= x<sub>i </sub>&lt; y<sub>i</sub> &lt;= 500</code></li>
	<li>The elements in <code>pairs</code> are unique.</li>
</ul>

#define printf(...) (0)

void printContainer(const auto &container) {
#ifndef printf(...)
    for (int x : container) {
        printf("%d ", x);
    }
    printf("\n");
#endif
}

class Solution {
    unordered_map<int, unordered_set<int>> adj;

    int getRoot(const unordered_set<int>& nodes) {
        printf("getRoot: ");
        printContainer(nodes);
        for (int u : nodes) {
            int cnt = 0;
            for (int v : adj[u]) {
                if (nodes.count(v)) {
                    ++cnt;
                }
            }
            if (cnt == nodes.size() - 1) {
                printf("root: %d\n\n", u);
                return u;
            }
        }
        printf("no root found\n\n");
        return -1;
    }

    vector<unordered_set<int>> getConnectedComponents(const unordered_set<int>& nodes) {
        vector<unordered_set<int>> ans;
        unordered_set<int> seen;
        for (int s : nodes) {
            if (seen.count(s)) {
                continue;
            }
            ans.emplace_back();
            function<void(int)> dfs;
            dfs = [&](int u) {
                seen.insert(u);
                ans.back().insert(u);
                for (int v : adj[u]) {
                    if (nodes.count(v) && seen.count(v) == 0) {
                        dfs(v);
                    }
                }
            };
            dfs(s);
        }
        return ans;
    }

public:
    int checkWays(vector<vector<int>>& pairs) {
        for (const auto& pair : pairs) {
            adj[pair[0]].insert(pair[1]);
            adj[pair[1]].insert(pair[0]);
        }

        queue<unordered_set<int>> todo;
        todo.emplace();
        for (auto& [u, neighbors] : adj) {
            todo.back().insert(u);
        }
        bool more_than_one_ways = false;
        while (!todo.empty()) {
            auto nodes = move(todo.front());
            todo.pop();

            int root = getRoot(nodes);
            if (root == -1) {
                return 0;
            }

            nodes.erase(root);
            for (int v : adj[root]) {
                adj[v].erase(root);
            }

            vector<unordered_set<int>> componenents = getConnectedComponents(nodes);
            printf("root: %d\ncomponenents.size(): %d\n\n", root, componenents.size());
            if (componenents.size() == 1) {
                more_than_one_ways = true;
            }
            for (auto& componenent : componenents) {
                todo.emplace(move(componenent));
            }
        }
        return more_than_one_ways ? 2 : 1;
    }
};
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