并查集模板 + 经典题目

本文参照AcWing cornerCao同学的讲解，整理了一下并查集的模板和3道经典题目的实现。

并查集模板 

class UnionFind{
public:
    vector father;
    UnionFind(int num){ //num表示元素的个数
        for(int i = 0; i < num; i++){
            father.push_back(i); //箭头指向自己
        }
    }
    int Find(int n){
        //递归
        if(father[n] == n)
            return n;
        father[n] = Find(father[n]); //路径压缩版本
        return father[n];
    }
    void Union(int a, int b){
        int fa = Find(a);
        int fb = Find(b);
        father[fb] = fa;
    }
};

经典题目 1：leetcode 547. 朋友圈

class UnionFind{
public:
    vector father;
    UnionFind(int num){
        for(int i = 0; i < num; i++){
            father.push_back(i);
        }
    }
    int Find(int n){
        if(father[n] == n)
            return n;
        father[n] = Find(father[n]);
        return father[n];
    }
    void Union(int a, int b){
        int fa = Find(a);
        int fb = Find(b);
        father[fb] = fa;
    }
};

class Solution {
public:
    int findCircleNum(vector>& M) {
        int N = M.size();
        UnionFind UF(N);
        for(int i = 0; i < N; i++){
            for(int j = 0; j < N; j++){
                if(M[i][j]){
                    UF.Union(i, j);
                }
            }
        }
        int res = 0;
        for(int i = 0; i < N; i++){
            if(UF.Find(i) == i)
                res++;
        }
        return res;
    }
};

经典题目 2：leetcode 200. 岛屿的个数

class UnionFind{
public:
    vector father;
    UnionFind(int num){
        for(int i = 0; i < num; i++){
            father.push_back(i);
        }
    }
    int Find(int n){
        if(father[n] == n)
            return n;
        father[n] = Find(father[n]);
        return father[n];
    }
    void Union(int a, int b){
        int fa = Find(a);
        int fb = Find(b);
        father[fb] = fa;
    }
};

class Solution {
public:
    int encode(int i, int j, int n){
        return i * n + j;
    }
    
    int numIslands(vector>& grid) {
        int M = grid.size();
        if(M == 0)
            return 0;
        int N = grid[0].size();
        if(N == 0)
            return 0;
        int directions[4][2] = {{1, 0}, {-1, 0}, {0, 1}, {0, -1}};
        UnionFind UF(M * N);
        for(int i = 0; i < M; i++){
            for(int j = 0; j < N; j++){
                if(grid[i][j] == '1'){
                    for(int d = 0; d < 4; d++){
                        int di = directions[d][0];
                        int dj = directions[d][1];
                        if(i + di >= 0 && i + di < M && j + dj >= 0 && j + dj < N && grid[i+di][j+dj] == '1'){
                            UF.Union(encode(i, j, N), encode(i + di, j + dj, N));
                        }
                    }
                }
            }
        }
        int res = 0;
        for(int i = 0; i < M; i++){
            for(int j = 0; j < N; j++){
                if(grid[i][j] == '1' && UF.Find(encode(i, j, N)) == encode(i, j, N))
                    res++;
            }
        }
        return res;
    }
};

经典题目 3：leetcode 684. 冗余连接

class UnionFind{
public:
    vector father;
    UnionFind(int num){
        for(int i = 0; i < num; i++){
            father.push_back(i);
        }
    }
    int Find(int n){
        if(father[n] == n)
            return n;
        father[n] = Find(father[n]);
        return father[n];
    }
    bool Union(int a, int b){
        int fa = Find(a);
        int fb = Find(b);
        father[fb] = fa;
        return fa == fb;
    }
};

class Solution {
public:
    vector findRedundantConnection(vector>& edges) {
        int N = edges.size();
        UnionFind UF(N + 1);
        vector res(2, 0);
        for(int i = 0; i < edges.size(); i++){
            int u = edges[i][0];
            int v = edges[i][1];
            if(UF.Union(u, v)){
                res[0] = u;
                res[1] = v;
            }
        }
        return res;
    }
};