Leetcode 2316. 统计无向图中无法互相到达点对数 DFS/BFS/并查集+前缀和

原题链接：Leetcode 2316. 统计无向图中无法互相到达点对数

DFS

class Solution {
public:
    vector<vector<int>> adj;
    vector<int> visit;
    void dfs(int i,int color)
    {
        visit[i]=color;
        for(auto x:adj[i])
        {
            if(!visit[x]) dfs(x,color);
        }
    }
    long long countPairs(int n, vector<vector<int>>& edges) {
        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);
        }
        long long color=0;
        map<int,long long> mp;
        for(int i=0;i<n;i++)
        {
            if(!visit[i]) 
            {
                color++;
                dfs(i,color);
                mp[color]=1;
            }
            else mp[visit[i]]++;
        }
        long long res=0,now=0;
        for(auto x: mp)
        {
            now+=x.second;
            res+=(x.second*(n-now));
        }
        return res;
    }
};

BFS:

class Solution {
public:
    vector<vector<int>> adj;
    vector<int> visit;
    long long countPairs(int n, vector<vector<int>>& edges) {
        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);
        }
        long long color=0;
        map<int,long long> mp;
        queue<int> q; 
        for(int i=0;i<n;i++)
        {
            if(!visit[i]) 
            {
                color++;
                q.push(i);
                visit[i]=1;
                while(!q.empty())
                {
                    mp[color]++;
                    int x=q.front(); q.pop();
                    for(auto y:adj[x]) 
                    {
                        if(!visit[y]) 
                        {
                            visit[y]=1;
                            q.push(y);
                        }
                    }
                }
            }
        }
        long long res=0,now=0;
        for(auto x: mp)
        {
            now+=x.second;
            res+=(x.second*(n-now));
        }
        return res;
    }
};

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(a!=x)
        {
            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 if(fx>fy) fa[fx]=fy;
        }
    }
    long long countPairs(int n, vector<vector<int>>& edges) {
        for(int i=0;i<n;i++) fa[i]=i;
        for(auto x:edges)
        {
            int a=x[0],b=x[1];
            Union(a,b);
        }
        map<int,long long> mp;
        for(int i=0;i<n;i++) 
        {
            int father=findfather(i);
            mp[father]++;
        }
        long long res=0,now=0;
        for(auto x: mp)
        {
            now+=x.second;
            res+=(x.second*(n-now));
        }
        return res;
    }
};