LeetCode684 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.

题意就是输入一组无向的边，然后依次往图中添加这些边，直到找到第一个使得这个图有环的边。
有一个解法就是保留每个顶点的出发点，若添加的边的两个顶点的出发点相同则会出现环，于是输出。具体做法就是初始化每个顶点的出发点为自身，然后加入边的时候更新每个顶点的出发点。复杂度为O(e*d)。

class Solution {
public:
    vector<int> findRedundantConnection(vector<vector<int>>& edges) {
        //int eSize = edges.size();
        vector<int> start(3000);
        //memset(start, 0, sizeof(int)*eSize);
        for (int i = 0; i < 3000; i++) {
            start[i] = i;
        }
        int tStart, tEnd;
        for (int i = 0; i < edges.size(); i++) {
            tStart = edges[i][0];
            tEnd = edges[i][1];

            if (start[tStart] == start[tEnd]) {
                return edges[i];
            }
            int needUpdate = start[tEnd];

            for (int j = 0; j < 3000; j++) {
                if (start[j] == needUpdate) {
                    start[j] = start[tStart];
                }
            }

        }

    }
};