Leetcode 2368. 受限条件下可到达节点的数目 BFS/DFS/并查集

原题链接：Leetcode 2368. 受限条件下可到达节点的数目

DFS

class Solution {
public:
    vector<vector<int>> adj;
    vector<int> visit;
    map<int,int> mp;
    int num=0;
    void dfs(int now)
    {
        visit[now]=1;
        num++;
        for(auto x:adj[now])
        {
            if(!visit[x] && !mp[x]) dfs(x);
        }
    }
    int reachableNodes(int n, vector<vector<int>>& edges, vector<int>& restricted) {
        adj.resize(n);
        visit.resize(n);
        for(auto x:edges)
        {
            int a=x[0],b=x[1];
            adj[a].push_back(b);
            adj[b].push_back(a);
        }
        for(auto x:restricted) mp[x]=1;
        dfs(0);
        return num;
    }
};
           

BFS

class Solution {
public:
    int reachableNodes(int n, vector<vector<int>>& edges, vector<int>& restricted) {
        vector<vector<int>> adj(n);
        vector<int> visit(n);
        map<int,int> mp;
        int num=0;
        for(auto x:edges)
        {
            int a=x[0],b=x[1];
            adj[a].push_back(b);
            adj[b].push_back(a);
        }
        for(auto x:restricted) mp[x]=1;
        queue<int> q;
        q.push(0);
        while(!q.empty())
        {
            int x=q.front(); q.pop();
            num++;
            visit[x]=1;
            for(auto y:adj[x])
            {
                if(!visit[y] && !mp[y])  q.push(y);
            }
        }
        return num;
    }
};
           

并查集

class Solution {
public:
    map<int,int> fa;
    int findfather(int x)
    {
        int a=x;
        while(fa[x]!=x) x=fa[x];
        while(fa[a]!=a)
        {
            int z=a;
            a=fa[a];
            fa[z]=x;
        }
        return x;
    }
    void Union(int x,int y)
    {
        int fx=findfather(x);
        int fy=findfather(y);
        if(fx!=fy)
        {
            if(fx<fy) fa[fy]=fx;
            else fa[fx]=fy;
        }
    }
    int reachableNodes(int n, vector<vector<int>>& edges, vector<int>& restricted) {
        map<int,int> mp;
        for(int i=0;i<n;i++) fa[i]=i;
        for(auto x:restricted) mp[x]=1;
        for(auto x:edges)
        {
            int a=x[0],b=x[1];
            if(!mp[a] && !mp[b]) Union(a,b);
        }
        int num=0;
        for(int i=0;i<n;i++)
        {
            if(mp[i]) continue;
            if(fa[i]==0) num++;
            else if(findfather(i)==0) num++;
        }
        return num;
    }
};