In this problem, a rooted tree is a directed graph such that, there is exactly one node (the root) for which all other nodes are descendants of this node, plus every node has exactly one parent, except for the root node which has no parents.

The given input is a directed graph that started as a rooted tree with N nodes (with distinct values 1, 2, …, N), with one additional directed edge added. The added edge has two different vertices chosen from 1 to N, and was not an edge that already existed.

The resulting graph is given as a 2D-array of edges. Each element of edges is a pair [u, v] that represents a directed edge connecting nodes u and v, where u is a parent of child v.

Return an edge that can be removed so that the resulting graph is a rooted tree of N nodes. If there are multiple answers, return the answer that occurs last in the given 2D-array.

Example 1:

Input: [[1,2], [1,3], [2,3]]
Output: [2,3]
Explanation: The given directed graph will be like this:
  1
 / \
v   v
2-->3

Example 2:

Input: [[1,2], [2,3], [3,4], [4,1], [1,5]]
Output: [4,1]
Explanation: The given directed graph will be like this:
5 <- 1 -> 2
     ^    |
     |    v
     4 <- 3
 

Note:

The size of the input 2D-array will be between 3 and 1000.

Every integer represented in the 2D-array will be between 1 and N, where N is the size of the input array.

解法1
從問題可以得知，要想構成一顆有向樹，必須滿足：

• 每個節點只有一個父節點
• 樹中沒有環（與無向圖做法一致）

特殊樣例

• [[1,2], [2,3], [3,1]] 每個節點有一個父節點但是有環
• [[1,2], [2,3], [1,3]] 節點3有兩個父節點，故刪除邊從[2,3]和[1,3]中選擇。
• [[4,2], [2,1], [1,4], [3, 1]] 節點1有兩個父節點。刪除[2,1]或者[3,1]，預設情況下刪除[3,1]，但由於[2,1]可以構成環，故需要刪除[2,1]

class Solution {
    int fa[1002];
public:
    vector<int> findRedundantDirectedConnection(vector<vector<int>>& edges) {
        vector<int> e1, e2; // 如果有某個節點有兩個父節點，則將這兩條邊存起來，刪除的話只會在這裡面刪除。
        memset(fa, -1, sizeof(fa)); // 只記錄節點的父節點，不做路徑壓縮
        for(auto e: edges) {
            int u = e[0], v = e[1];
            if(fa[v] != -1) {
                e1 = {fa[v], v};
                e2 = {u, v}; 
                break;
            }
            else {
                fa[v] = u; //v的父節點記錄為u
            }
        }
        memset(fa, -1, sizeof(fa));
        if(e1.empty()) { // 找不到某個結點有兩個父節點，必然存在環。
            for(auto e: edges) {
                int u=e[0], v= e[1];
                int fu=find(u), fv=find(v);
                if(fu!=fv) {
                    fa[fu]=fv;
                } 
                else {
                    return {u, v}; // 跟有向圖的判斷類似
                }
            }
        }
        else {
            for(auto e: edges) {
                int u=e[0], v=e[1];
                if(u==e2[0]&&v==e2[1]) continue;
                int fu=find(u), fv=find(v);
                if(fu!=fv) {
                    fa[fu]=fv;
                }
                else 
                    return e1;
            }
        }
        return e2;
    }
    

    int find(int x) {
        return fa[x]==-1?x:fa[x]=find(fa[x]);
    }
};