Redundant Connection

In this problem, a tree is an undirected graph that is connected and has no cycles.

The given input is a graph that started as a tree with N nodes (with distinct values 1, 2, …, N), with one additional 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] with u < v, that represents an undirected edge connecting nodes u and v.

Return an edge that can be removed so that the resulting graph is a tree of N nodes. If there are multiple answers, return the answer that occurs last in the given 2D-array. The answer edge [u, v] should be in the same format, with u < v.

Example 1:

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

Example 2:

Input: [[1,2], [2,3], [3,4], [1,4], [1,5]]Output: [1,4]Explanation: The given undirected graph will be like this:5 - 1 - 2    |   |    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.

FROM:
LeetCode-Algorithms-graph

Hint:
这道题说是要把图上的一条边删去，把无向图变成一颗树。但事实上，从题目最后说“如果有多种选择选择后出现的一种”可以知道，其实是要一条条边加进去，去画一棵树。那这样我们每次新加一条边，就要判断一下，会不会构成环路。如果构成回路，就说明这条新边就是我们要删除的边。
那么，怎么判断是否构成回路呢？构成回路的条件是很显然的，新加入的边所对应的两个节点中间有通路。当然如果要去遍历树看看是否有通路，这样的开销太大了。我们试着想一下，在加入这条边之前，这个无向图是一个森林或者一棵树，如果两个节点之间有通路，那么它们必然在一棵树上，那么它们必然有一个共同的祖先，即树根。
这样我们只要维护一个数组，记录节点的祖先节点，每次加入新边时，判断两个节点的祖先节点是否相同即可。当然我们判断祖先是否相同时，一定要判断节点辈分最高的祖先。

class Solution {public:    vector<int> findRedundantConnection(vector<vector<int>>& edges) {        int *parent = new int[edges.size()+1];        for (int i = 0; i < edges.size()+1; i++) {            parent[i] = i;        }        int p1, p2;        for (auto e : edges) {            p1 = getParent(e[0], parent);            p2 = getParent(e[1], parent);            if (p1 == p2) {                return e;            } else {                parent[p2] = p1;            }        }        vector<int> no;        return no;    }    int getParent(int a, int *parent) {        if (a != parent[a]) {            parent[a] = getParent(parent[a], parent);        }        return parent[a];    }};