技術標籤：leetcode每日一題202101leetcode並查集

Problem

There are n computers numbered from 0 to n-1 connected by ethernet cables connections forming a network where connections[i] = [a, b] represents a connection between computers a and b. Any computer can reach any other computer directly or indirectly through the network.

Given an initial computer network connections. You can extract certain cables between two directly connected computers, and place them between any pair of disconnected computers to make them directly connected. Return the minimum number of times you need to do this in order to make all the computers connected. If it’s not possible, return -1.

Constraints:

• 1 <= n <= 10^5
• 1 <= connections.length <= min(n*(n-1)/2, 10^5)
• connections[i].length == 2
• 0 <= connections[i][0], connections[i][1] < n
• connections[i][0] != connections[i][1]
• There are no repeated connections.
• No two computers are connected by more than one cable.

Example1

Input: n = 4, connections = [[0,1],[0,2],[1,2]]

Output: 1
Explanation: Remove cable between computer 1 and 2 and place between computers 1 and 3.

Example2

Input: n = 6, connections = [[0,1],[0,2],[0,3],[1,2],[1,3]]
Output: 2

Example3

Input: n = 6, connections = [[0,1],[0,2],[0,3],[1,2]]
Output: -1
Explanation: There are not enough cables.

Example4

Input: n = 5, connections = [[0,1],[0,2],[3,4],[2,3]]
Output: 0

Solution

// 並查集模板
class UnionFind {
public:
    vector<int> parent;
    vector<int> size;
    int n;
    // 當前連通分量數目
    int setCount;
    
public:
    UnionFind(int _n): n(_n), setCount(_n), parent(_n), size(_n, 1) {
        iota(parent.begin(), parent.end(), 0);
    }
    
    int findset(int x) {
        return parent[x] == x ? x : parent[x] = findset(parent[x]);
    }
    
    bool unite(int x, int y) {
        x = findset(x);
        y = findset(y);
        if (x == y) {
            return false;
        }
        if (size[x] < size[y]) {
            swap(x, y);
        }
        parent[y] = x;
        size[x] += size[y];
        --setCount;
        return true;
    }
    
    bool connected(int x, int y) {
        x = findset(x);
        y = findset(y);
        return x == y;
    }
};

class Solution {
public:
    int makeConnected(int n, vector<vector<int>>& connections) {
        if (connections.size() < n - 1) {
            return -1;
        }

        UnionFind uf(n);
        for (const auto& conn: connections) {
            uf.unite(conn[0], conn[1]);
        }

        return uf.setCount - 1;
    }
};

//作者：LeetCode-Solution
//連結：https://leetcode-cn.com/problems/number-of-operations-to-make-network-connected/solution/lian-tong-wang-luo-de-cao-zuo-ci-shu-by-leetcode-s/