886. Possible Bipartition ---M

Given a set of N people (numbered 1, 2, ..., N), we would like to split everyone into two groups of any size.

Each person may dislike some other people, and they should not go into the same group. 

Formally, if dislikes[i] = [a, b], it means it is not allowed to put the people numbered a and b into the same group.

Return true if and only if it is possible to split everyone into two groups in this way.

 

Example 1:

Input: N = 4, dislikes = [[1,2],[1,3],[2,4]]
Output: true
Explanation: group1 [1,4], group2 [2,3]

Example 2:

Input: N = 3, dislikes = [[1,2],[1,3],[2,3]]
Output: false

Example 3:

Input: N = 5, dislikes = [[1,2],[2,3],[3,4],[4,5],[1,5]]
Output: false

 

Constraints:

• 1 <= N <= 2000
• 0 <= dislikes.length <= 10000
• dislikes[i].length == 2
• 1 <= dislikes[i][j] <= N
• dislikes[i][0] < dislikes[i][1]
• There does not exist i != j for which dislikes[i] == dislikes[j].

 A:

----------BFS---------  用queue

class Solution {
public:
    bool possibleBipartition(int N, vector<vector<int>>& dislikes) {
        vector<int> V(N+1, -1); // -1 undecided, 0, one color, 1 another
        unordered_map<int, vector<int>> M;
        for(auto dis : dislikes){            
            int a = dis[0];
            int b = dis[1];
            M[a].push_back(b);
            M[b].push_back(a);
        }
        
        for(int i =1;i<=N; i++){
            if(V[i] >=0 )
                continue;
            queue<int> Q;
            V[i] = 0;
            Q.push(i);
            while(!Q.empty()){
               int a = Q.front();
                Q.pop();
                for(auto end : M[a]){
                    if(V[end]==V[a]){
                        return false;
                    }else if(V[end]==-1){
                        V[end] = 1-V[a];
                        Q.push(end);
                    }
                }
            }
        }
        return true;
    }
};