无向图中连通分量的个数

1、DFS
每调用一次DFS(u)，就能遍历u所在的连通分量

class Solution {
private:
    static const int N = 2010;
    int n, G[N][N];
    bool visited[N] = {false};
public:
    int countComponents(int n, vector>& edges) {
        this->n = n;
        memset(G, 0, sizeof G);
        for (vector edge : edges){
            int u = edge[0], v = edge[1];
            G[u][v] = G[v][u] = 1;
        }
        int res = 0;
        for (int i = 0; i < n; i++){
            if (!visited[i]){
                DFS(i);
                res++;
            }
        }
        return res;
    }

    void DFS(int u){
        visited[u] = true;
        for (int v = 0; v < n; v++){
            if (G[u][v] != 0 && !visited[v])
                DFS(v);
        }
    }
};

2、并查集
将两个节点看做集合，如果a、b之间有边，那么把a、b合并，最后计算集合中根节点的个数即可

class Solution {
private:
    static const int N = 2010;
    int father[N];
public:
    int countComponents(int n, vector>& edges) {
        int res = 0;
        memset(father, 0, sizeof father);
        for (int i = 0; i < n; i++) father[i] = i;
        for (vector edge : edges){
            int a = edge[0], b = edge[1];
            int FaA = findFather(a), FaB = findFather(b);
            if (FaA != FaB){
                father[FaB] = FaA;  //将b的父节点连接到a所在的子树上
            }
        }
        for (int i = 0; i < n; i++){
            if (father[i] == i) res++;
        }
        return res;
    }

    int findFather(int a){
        if (a != father[a]) father[a] = findFather(father[a]);
        return father[a]; 
    }
};