2360. Longest Cycle in a Graph

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

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

Return the length of the longest cycle in the graph. If no cycle exists, return -1.

A cycle is a path that starts and ends at the same node.

Example 1:

Input: edges = [3,3,4,2,3]
Output: 3

Explanation: The longest cycle in the graph is the cycle: 2 -> 4 -> 3 -> 2.
The length of this cycle is 3, so 3 is returned.

Example 2:

Input: edges = [2,-1,3,1]
Output: -1

Explanation: There are no cycles in this graph.

Constraints:

• n == edges.length
• 2 <= n <= 105
• -1 <= edges[i] < n
• edges[i] != i

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每一个 node 可能有多个 from， 但是 to 只有一个， 所以我们将每个 node 的 to 收集起来作为当前 node 的 parents， 同时记录下 node[i]与 parents[i]的距离， 我们 bfs 更新 parents， 假设 parents[i] = (parent, parent_distance), parents[parent] = (parent_of_parent, parent_of_parent_distance), 那我们就可以更新 parents[i] = (parent_of_parent, parent_distance + parent_of_parent_distance), 但是要注意， 如果 parent == parent_of_parent 则证明 parents[parent]已经形成环路了， 我们就没必要再更新当前 parents[i]了, 因为此时 parent_distance + parent_of_parent_distance 是当前节点到环路的入口节点之间的距离加上环路的长度， 对我们的答案没有任何用处。如果 parent_of_parent == i, 则证明当前节点形成环路了， 此时的 parent_distance + parent_of_parent_distance 就是环路的长度， 我们只需要收集该值的最大值就是最终答案了

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impl Solution {
    pub fn longest_cycle(edges: Vec<i32>) -> i32 {
        let mut parents: Vec<(i32, i32)> = edges.into_iter().map(|v| (v, 1)).collect();
        let mut ans = 0;
        loop {
            let mut changed = false;
            for i in 0..parents.len() {
                if parents[i].0 < 0 {
                    continue;
                }
                let (p, pn) = parents[i];
                let (pp, ppn) = parents[p as usize];
                if p == pp {
                    continue;
                }
                parents[i] = (pp, pn + ppn);
                changed = true;
                if pp == i as i32 {
                    ans = ans.max(pn + ppn);
                }
            }
            if !changed {
                break;
            }
        }
        if ans == 0 {
            -1
        } else {
            ans
        }
    }
}