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package LeetPlan;
import java.util.ArrayList;
import java.util.List;
import java.util.Map;
import java.util.Set;
/**
*
* 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.
*
* @author Liu.3502
* @created 2018-05-04 下午3:08
*/
public class LC685_RedundantConnectionII {
public static void main(String[] args) {
int s = 2;
//[1, 3, 2, 8, 4, 9]
List<List<String>> num = new ArrayList<>();
List<List<String>> ans1 = accountsMergeBF(num);
List<List<String>> ans2 = accountsMerge(num);
System.out.println( ans1 + " / " + ans2);
}
// we store all the account in to set one by one and check the new input account is in the old set or not
public static List<List<String>> accountsMergeBF(List<List<String>> accounts) {
List<Map<Set<String>,String>> allNode = new ArrayList<>();
// for(){
//
// }
return null;
}
public static List<List<String>> accountsMerge(List<List<String>> accounts) {
return null;
}
}