886.Possible Bipartition

Tags: Medium Depth-first Search

Links: https://leetcode.com/problems/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. 1 <= N <= 2000
2. 0 <= dislikes.length <= 10000
3. 1 <= dislikes[i][j] <= N
4. dislikes[i][0] < dislikes[i][1]
5. There does not exist i != j for which dislikes[i] == dislikes[j].

---

class Solution {
    vector<vector<int>> denseGraph;
    vector<int> color;
    int n;
public:
    bool possibleBipartition(int N, vector<vector<int>>& dislikes) {
        std::ios_base::sync_with_stdio(false);
        cin.tie(NULL);
        cout.tie(NULL);

        denseGraph.resize(N + 1, vector<int>(N + 1, 0));
        color.resize(N + 1, -1);
        n = N;

        for (auto & e : dislikes) {
            denseGraph[e[0]][e[1]] = 1;
            denseGraph[e[1]][e[0]] = 1;
        }

        bool flag = true;
        for (int i = 1; i <= N; ++i) {
            if (color[i] == -1) { //第i个点还没有被染色
                if (!DFS(i, 0)) { flag = false; break; }
            }
        }

        return flag;
    }

    bool DFS(int v, int c)
    {
        color[v] = (c + 1) % 2;
        for (int i = 1; i <= n; ++i) {
            if (denseGraph[v][i]) {
                if (color[i] == -1) {
                    if (!DFS(i, color[v])) return false;
                }
                else if (color[i] != c) return false;
            }
        }
        return true;
    }
};

二部图染色问题。