leetcode 890. Possible Bipartition

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

Note:

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

思路：二分图染色

class Solution {
public:
    vector<int>G[2100];
    int color[2100];
    //memset(color, -1, sizeof (color));
    int bfs() {
        queue<int>q;
        q.push(1);
        color[1] = 1;
        while(!q.empty()) {
            int v1 = q.front();
            q.pop();
            for(int i = 0; i < G[v1].size(); i++) {
                int v2 = G[v1][i];
                if(color[v2] == -1) {
                    color[v2] = -color[v1];
                    q.push(v2);
                }
                else if(color[v1] == color[v2])
                    return 0;
            }
        }
        return 1;
    }
    bool possibleBipartition(int N, vector<vector<int>>& dislikes) {
        for (int i = 0; i < dislikes.size(); ++i) {
            vector<int> x = dislikes[i];
            //cout << x[0]  << " " <<x[1] << endl;
            G[x[0]].push_back(x[1]);
            G[x[1]].push_back(x[0]);
        }
        for (int i = 0; i <= 2000; ++i)color[i] = -1;
        return bfs();
    }
};

原文地址：https://www.cnblogs.com/pk28/p/9462393.html