LeetCode2092. Find All People With Secret——并查集

一、题目

You are given an integer n indicating there are n people numbered from 0 to n - 1. You are also given a 0-indexed 2D integer array meetings where meetings[i] = [xi, yi, timei] indicates that person xi and person yi have a meeting at timei. A person may attend multiple meetings at the same time. Finally, you are given an integer firstPerson.

Person 0 has a secret and initially shares the secret with a person firstPerson at time 0. This secret is then shared every time a meeting takes place with a person that has the secret. More formally, for every meeting, if a person xi has the secret at timei, then they will share the secret with person yi, and vice versa.

The secrets are shared instantaneously. That is, a person may receive the secret and share it with people in other meetings within the same time frame.

Return a list of all the people that have the secret after all the meetings have taken place. You may return the answer in any order.

Example 1:

Input: n = 6, meetings = [[1,2,5],[2,3,8],[1,5,10]], firstPerson = 1
Output: [0,1,2,3,5]
Explanation:
At time 0, person 0 shares the secret with person 1.
At time 5, person 1 shares the secret with person 2.
At time 8, person 2 shares the secret with person 3.
At time 10, person 1 shares the secret with person 5.​​​​
Thus, people 0, 1, 2, 3, and 5 know the secret after all the meetings.
Example 2:

Input: n = 4, meetings = [[3,1,3],[1,2,2],[0,3,3]], firstPerson = 3
Output: [0,1,3]
Explanation:
At time 0, person 0 shares the secret with person 3.
At time 2, neither person 1 nor person 2 know the secret.
At time 3, person 3 shares the secret with person 0 and person 1.
Thus, people 0, 1, and 3 know the secret after all the meetings.
Example 3:

Input: n = 5, meetings = [[3,4,2],[1,2,1],[2,3,1]], firstPerson = 1
Output: [0,1,2,3,4]
Explanation:
At time 0, person 0 shares the secret with person 1.
At time 1, person 1 shares the secret with person 2, and person 2 shares the secret with person 3.
Note that person 2 can share the secret at the same time as receiving it.
At time 2, person 3 shares the secret with person 4.
Thus, people 0, 1, 2, 3, and 4 know the secret after all the meetings.

Constraints:

2 <= n <= 105
1 <= meetings.length <= 105
meetings[i].length == 3
0 <= xi, yi <= n - 1
xi != yi
1 <= timei <= 105
1 <= firstPerson <= n - 1

二、题解

class Solution {
public:
    int father[100001];
    bool secret[100001];
    static bool cmp(vector<int>& a,vector<int>& b){
        return a[2] < b[2]; 
    }
    vector<int> findAllPeople(int n, vector<vector<int>>& meetings, int firstPerson) {
        vector<int> res;
        int m = meetings.size();
        build(n,firstPerson);
        sort(meetings.begin(),meetings.end(),cmp);
        for(int l = 0;l < m;){
            int r = l;
            while(r + 1 < m && meetings[l][2] == meetings[r+1][2]) r++;
            for(int i = l;i <= r;i++){
                unions(meetings[i][0],meetings[i][1]);
            }
            for(int i = l;i <= r;i++){
                int a = meetings[i][0],b = meetings[i][1];
                if(!secret[find(a)]) father[a] = a;
                if(!secret[find(b)]) father[b] = b;
            }
            l = r + 1;
        }
        for(int i = 0;i < n;i++){
            if(secret[find(i)]) res.push_back(i);
        }
        return res;
    }
    void build(int n,int first){
        for(int i = 0;i < n;i++){
            father[i] = i;
            secret[i] = false;
        }
        father[first] = 0;
        secret[0] = true;
    }
    int find(int x){
        if(x != father[x]) father[x] = find(father[x]);
        return father[x];
    }
    void unions(int x,int y){
        int fx = find(x),fy = find(y);
        if(fx != fy){
            father[fx] = fy;
            secret[fy] |= secret[fx];
        }
    }
};