684. Redundant Connection

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问题描述

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.

Update (2017-09-26):
We have overhauled the problem description + test cases and specified clearly the graph is an undirected graph. For the directed graph follow up please see Redundant Connection II). We apologize for any inconvenience caused.

题意理解

N个结点(记作1~N)的一棵树,加上一条边之后变成一个无向图. 求一条边, 删除后可以使图变成一棵树.(如果有多条符合条件的边, 返回在edges数组中的最后一条)

算法分析

由于这个图原来是一棵树, 也就是没有环存在, 所以再加一条边只会构成一个环.
做union-find, 在union过程中, 找到最后一条满足:边的两个顶点指向同一个祖先的边, 即为解

具体实现

class Solution {public:    vector<int> findRedundantConnection(vector<vector<int>>& edges) {        vector<int> root(edges.size()+1);        vector<int> res;        if (edges.empty()) return res;        //initialize        for (int i = 1; i <= edges.size(); i++)            root[i] = i;        // add edge        for (int i = 0; i < edges.size(); i++) {            myUnion(edges[i][0], edges[i][1], root, res);        }        return res;    }    void myUnion(int p, int q, vector<int>& root, vector<int>& res) {        int proot = find(p, root);        int qroot = find(q, root);        if (proot != qroot) {            root[qroot] = proot;        }        else {            res.clear();            res.push_back(p);            res.push_back(q);        }    }    int find(int p, vector<int>& root) {        while (p != root[p])            p = find(root[p], root);        return p;    }};