求解朋友关系中的朋友圈数量

Posted 2020-09-16 赵弘添

问题描述：给出10w条人和人之间的朋友关系，求出这些朋友关系中有多少个朋友圈

样例A-B、B-C、D-E、E-F ，这四对关系中存在2个朋友圈

解题思路：并查集，而题目只需要求出朋友圈数量，并不需要求出各朋友圈，所以该并查集的实现也可以非常简单。

A-B,就把father[B] = A，处理每条朋友关系即可得到结果。

而关于并查集的介绍，已有很多博文有所阐述，这里就不啰嗦了。

如下给出实现的并查集

Python实现

class WeightedUF():  
    fatherid=[]  
    sz=[]  
    count=0  
    def __init__(self,n):  
        self.count=n  
        self.fatherid=[i for i in range(n)]  
        self.sz=[0 for i in range(n)]  
    def getcount(self):  
        return self.count  
    def connected(self,p,q):  
        return self.find(p)==self.find(q)  
    def find(self,p):  
        while p !=self.fatherid[p]:  
            p=self.fatherid[p]  
        return p  
    def pathcompressionfind(self,p):  
        if p==self.fatherid[p]:  
            return p  
        else:  
            self.fatherid[p]=self.pathcompressionfind(self.fatherid[p])  
            return self.fatherid[p]  
    def union(self,p,q):  
        i=self.find(p)  
        j=self.find(q)  
        if i==j:  
            return   
        if self.sz[i]<self.sz[j]:  
            self.fatherid[i]=j  
            self.sz[j]+=self.sz[i]  
        else:  
            self.fatherid[j]=i  
            self.sz[i]+=self.sz[j]  
        self.count-=1  

Java实现

public class WeightUF {
    int[] fatherid ;
    int[] sz;
    int count = 0;
    public WeightUF(int n){
        this.count = n;
        this.fatherid = new int[n];
        this.sz = new int[n];
        for(int i=0;i<n;i++){
            fatherid[i] = i;
            sz[i] = 0;
        }
    }
    public int getCount(){
        return count;
    }
    public boolean connected(int p,int q){
        return find(p) == find(q);
    }
    public int find(int p){
        while (p != fatherid[p]){
            p = fatherid[p];
        }
        return p;
    }
    public int pathcompressionfind(int p){
        if(p == fatherid[p]){
            return p;
        }
        else{
            fatherid[p] = pathcompressionfind(p);
            return fatherid[p];
        }
    }
    public void union(int p,int q){
        int i = find(p);
        int j = find(q);
        if(i == j){
            return;
        }
        if(sz[i] < sz[j]){
            fatherid[i] = j;
            sz[j] += sz[i];
        }
        else{
            fatherid[j] = i;
            sz[i] += sz[j];
        }
        count -= 1;
    }
}

测试样例（java）

public static void main(String[] args) {
        WeightUF weightUF = new WeightUF(10);
        weightUF.union(9,2);
        weightUF.union(9,3);
        weightUF.union(1,2);
        weightUF.union(5,4);
        System.out.println(weightUF.getCount());
        System.out.println(weightUF.connected(9,4));
        System.out.println(weightUF.connected(9,5));
    }