<p>You are given a <strong>directed</strong> graph of <code>n</code> nodes numbered from <code>0</code> to <code>n - 1</code>, where each node has <strong>at most one</strong> outgoing edge.</p>

<p>The graph is represented with a given <strong>0-indexed</strong> array <code>edges</code> of size <code>n</code>, indicating that there is a directed edge from node <code>i</code> to node <code>edges[i]</code>. If there is no outgoing edge from node <code>i</code>, then <code>edges[i] == -1</code>.</p>

<p>Return <em>the length of the <strong>longest</strong> cycle in the graph</em>. If no cycle exists, return <code>-1</code>.</p>

<p>A cycle is a path that starts and ends at the <strong>same</strong> node.</p>

<p>&nbsp;</p>
<p><strong class="example">Example 1:</strong></p>
<img alt="" src="https://assets.leetcode.com/uploads/2022/06/08/graph4drawio-5.png" style="width: 335px; height: 191px;" />
<pre>
<strong>Input:</strong> edges = [3,3,4,2,3]
<strong>Output:</strong> 3
<strong>Explanation:</strong> The longest cycle in the graph is the cycle: 2 -&gt; 4 -&gt; 3 -&gt; 2.
The length of this cycle is 3, so 3 is returned.
</pre>

<p><strong class="example">Example 2:</strong></p>
<img alt="" src="https://assets.leetcode.com/uploads/2022/06/07/graph4drawio-1.png" style="width: 171px; height: 161px;" />
<pre>
<strong>Input:</strong> edges = [2,-1,3,1]
<strong>Output:</strong> -1
<strong>Explanation:</strong> There are no cycles in this graph.
</pre>

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

<ul>
	<li><code>n == edges.length</code></li>
	<li><code>2 &lt;= n &lt;= 10<sup>5</sup></code></li>
	<li><code>-1 &lt;= edges[i] &lt; n</code></li>
	<li><code>edges[i] != i</code></li>
</ul>

class Solution {
public:
    stack<int> st; int n, cnt = 0;
    int max_cycle = 0;
    vector<int> adj[100001]; 
    vector<int> t_adj[100001];// transpose graph 
    int visited[100001];

    // Sắp xếp topo
    void DFS1(int u){
        visited[u] = true;
        for(int v : adj[u]){
            if(!visited[v]){
                DFS1(v);
            }
        }
        st.push(u);
    }
    
    void DFS2(int u){
        visited[u] = true;
        cnt++;
        for(int v : t_adj[u]){
            if(!visited[v]){
                DFS2(v);
            }
        }
    }

    void kosarazu(){
        for(int i = 0; i < n; i++){
            if(!visited[i]) DFS1(i);
        }
        memset(visited, 0, sizeof(visited));
        while(!st.empty()){
            int top = st.top(); st.pop();
            if(!visited[top]){
                DFS2(top); 
                max_cycle = max(max_cycle, cnt);
            }
            cnt = 0;
        }
    }
    int longestCycle(vector<int>& edges) {
        n = edges.size();
        for(int i = 0; i < n; i++){
            if(edges[i] != -1){
                adj[i].push_back(edges[i]);
                t_adj[edges[i]].push_back(i);
            }
        }
        kosarazu();
        if(max_cycle <= 1) return -1;
        else return max_cycle;
    }
};
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