<p>There are <code>n</code> cities numbered from <code>0</code> to <code>n - 1</code> and <code>n - 1</code> roads such that there is only one way to travel between two different cities (this network form a tree). Last year, The ministry of transport decided to orient the roads in one direction because they are too narrow.</p>

<p>Roads are represented by <code>connections</code> where <code>connections[i] = [a<sub>i</sub>, b<sub>i</sub>]</code> represents a road from city <code>a<sub>i</sub></code> to city <code>b<sub>i</sub></code>.</p>

<p>This year, there will be a big event in the capital (city <code>0</code>), and many people want to travel to this city.</p>

<p>Your task consists of reorienting some roads such that each city can visit the city <code>0</code>. Return the <strong>minimum</strong> number of edges changed.</p>

<p>It&#39;s <strong>guaranteed</strong> that each city can reach city <code>0</code> after reorder.</p>

<p>&nbsp;</p>
<p><strong class="example">Example 1:</strong></p>
<img alt="" src="https://assets.leetcode.com/uploads/2020/05/13/sample_1_1819.png" style="width: 311px; height: 189px;" />
<pre>
<strong>Input:</strong> n = 6, connections = [[0,1],[1,3],[2,3],[4,0],[4,5]]
<strong>Output:</strong> 3
<strong>Explanation: </strong>Change the direction of edges show in red such that each node can reach the node 0 (capital).
</pre>

<p><strong class="example">Example 2:</strong></p>
<img alt="" src="https://assets.leetcode.com/uploads/2020/05/13/sample_2_1819.png" style="width: 509px; height: 79px;" />
<pre>
<strong>Input:</strong> n = 5, connections = [[1,0],[1,2],[3,2],[3,4]]
<strong>Output:</strong> 2
<strong>Explanation: </strong>Change the direction of edges show in red such that each node can reach the node 0 (capital).
</pre>

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

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

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

<ul>
	<li><code>2 &lt;= n &lt;= 5 * 10<sup>4</sup></code></li>
	<li><code>connections.length == n - 1</code></li>
	<li><code>connections[i].length == 2</code></li>
	<li><code>0 &lt;= a<sub>i</sub>, b<sub>i</sub> &lt;= n - 1</code></li>
	<li><code>a<sub>i</sub> != b<sub>i</sub></code></li>
</ul>

#pragma GCC optimize("Ofast,no-stack-protector")
#pragma GCC optimize("no-math-errno,unroll-loops")
#pragma GCC target("sse,sse2,sse3,ssse3,sse4")

static int speedup = []() {
    ios_base::sync_with_stdio(false);
    cin.tie(nullptr);
    cout.tie(nullptr);

    constexpr size_t cacheLineSize = 64; // Common cache line size in bytes
    constexpr size_t intSize = sizeof(int); // Size of an int in bytes
    constexpr size_t firstFew = cacheLineSize / intSize; // Number of ints that fit into a cache line

    std::cout << "Number of integers that fit into a cache line: " << firstFew << std::endl;

    return 0;
}();
class Solution {
public:
    //if an edge is outgoing from current location, we will have to flip it :(. therefore res++ in this case.
    int dfs(int s, vector<vector<int>>& outgoing, vector<vector<int>>& incoming, int parent){
        int res = 0;
        for(int i : outgoing[s]){
            if(parent == i) continue;
            res++;
            res += dfs(i, outgoing, incoming, s);
        }
        for(int i : incoming[s]){
            if(parent == i)continue;
            res += dfs(i, outgoing, incoming, s);
        }
        return res;
    }
    int minReorder(int n, vector<vector<int>>& connections) {
        vector<vector<int>> outgoing(n);
        vector<vector<int>> incoming(n);
        //because we are traversing forward and back along edges, it's important not to revisit visited locations.
        unordered_set<int> visited;
        for(auto& i : connections){
            outgoing[i[0]].push_back(i[1]);
            incoming[i[1]].push_back(i[0]);
        }

        return dfs(0, outgoing, incoming, -1);
    }
};
111716