[leetcode]886. Possible Bipartition

你看见我的代码了么

于 2019-03-03 11:13:30 发布

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分类专栏： LeetCode Medium DFS 文章标签： leetcode

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本文详细解析了LeetCode上编号为886的题目“Possible Bipartition”的解决方案。该问题要求将N个人分成两组，使得相互不喜欢的人不会被分在同一组。通过深度优先搜索(DFS)算法，实现了一个有效的解决方案来判断是否可以进行这样的分组。

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[leetcode]886. Possible Bipartition

Analysis

Still raining outside…—— [每天刷题并不难0.0]

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.
在这里插入图片描述

Explanation:

Solved by DFS,

group[i] = 0 means node i hasn’t been visited.
group[i] = 1 means node i has been grouped to 1.
group[i] = -1 means node i has been grouped to -1.

Implement

class Solution {
public:
    bool possibleBipartition(int N, vector<vector<int>>& dislikes) {
        vector<vector<int>> graph(N, vector<int>(N));
        for(vector<int> d:dislikes){
            graph[d[0]-1][d[1]-1] = 1;
            graph[d[1]-1][d[0]-1] = 1;
        }
        vector<int> group(N, 0);
        for(int i=0; i<N; i++){
            if(group[i]==0 && !DFS(N, graph, group, i, 1))
                return false;
        }
        return true;
    }
    bool DFS(int N, vector<vector<int>>& graph, vector<int>& group, int idx, int g){
        group[idx] = g;
        for(int i=0; i<N; i++){
            if(graph[idx][i] == 1){
                if(group[i] == g)
                    return false;
                if(group[i] == 0 && !DFS(N, graph, group, i, -g))
                    return false;
            }
        }
        return true;
    }
};