Shortest Path with Alternating Colors
<p>You are given an integer <code>n</code>, the number of nodes in a directed graph where the nodes are labeled from <code>0</code> to <code>n - 1</code>. Each edge is red or blue in this graph, and there could be self-edges and parallel edges.</p>

<p>You are given two arrays <code>redEdges</code> and <code>blueEdges</code> where:</p>

<ul>
	<li><code>redEdges[i] = [a<sub>i</sub>, b<sub>i</sub>]</code> indicates that there is a directed red edge from node <code>a<sub>i</sub></code> to node <code>b<sub>i</sub></code> in the graph, and</li>
	<li><code>blueEdges[j] = [u<sub>j</sub>, v<sub>j</sub>]</code> indicates that there is a directed blue edge from node <code>u<sub>j</sub></code> to node <code>v<sub>j</sub></code> in the graph.</li>
</ul>

<p>Return an array <code>answer</code> of length <code>n</code>, where each <code>answer[x]</code> is the length of the shortest path from node <code>0</code> to node <code>x</code> such that the edge colors alternate along the path, or <code>-1</code> if such a path does not exist.</p>

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

<pre>
<strong>Input:</strong> n = 3, redEdges = [[0,1],[1,2]], blueEdges = []
<strong>Output:</strong> [0,1,-1]
</pre>

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

<pre>
<strong>Input:</strong> n = 3, redEdges = [[0,1]], blueEdges = [[2,1]]
<strong>Output:</strong> [0,1,-1]
</pre>

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

<ul>
	<li><code>1 &lt;= n &lt;= 100</code></li>
	<li><code>0 &lt;= redEdges.length,&nbsp;blueEdges.length &lt;= 400</code></li>
	<li><code>redEdges[i].length == blueEdges[j].length == 2</code></li>
	<li><code>0 &lt;= a<sub>i</sub>, b<sub>i</sub>, u<sub>j</sub>, v<sub>j</sub> &lt; n</code></li>
</ul>
Breadth-First Search; Graph
class Solution {
public:
    vector<int> shortestAlternatingPaths(int n, vector<vector<int>>& redEdges, vector<vector<int>>& blueEdges) {
        vector<vector<vector<int>>> g(2, vector<vector<int>>(n));
        for (auto& e : redEdges) {
            g[0][e[0]].push_back(e[1]);
        }
        for (auto& e : blueEdges) {
            g[1][e[0]].push_back(e[1]);
        }
        queue<pair<int, int>> q;
        q.emplace(0, 0);
        q.emplace(0, 1);
        bool vis[n][2];
        memset(vis, false, sizeof vis);
        vector<int> ans(n, -1);
        int d = 0;
        while (!q.empty()) {
            for (int k = q.size(); k; --k) {
                auto [i, c] = q.front();
                q.pop();
                if (ans[i] == -1) {
                    ans[i] = d;
                }
                vis[i][c] = true;
                c ^= 1;
                for (int& j : g[c][i]) {
                    if (!vis[j][c]) {
                        q.emplace(j, c);
                    }
                }
            }
            ++d;
        }
        return ans;
    }
};