• 1107 Social Clusters (30 分)(并查集)


    When register on a social network, you are always asked to specify your hobbies in order to find some potential friends with the same hobbies. A "social cluster" is a set of people who have some of their hobbies in common. You are supposed to find all the clusters.

    Input Specification:

    Each input file contains one test case. For each test case, the first line contains a positive integer N (<=1000), the total number of people in a social network. Hence the people are numbered from 1 to N. Then N lines follow, each gives the hobby list of a person in the format:
    Ki: hi[1] hi[2] ... hi[Ki]
    where Ki (>0) is the number of hobbies, and hi[j] is the index of the j-th hobby, which is an integer in [1, 1000].

    Output Specification:

    For each case, print in one line the total number of clusters in the network. Then in the second line, print the numbers of people in the clusters in non-increasing order. The numbers must be separated by exactly one space, and there must be no extra space at the end of the line.

    Sample Input:

    8
    3: 2 7 10
    1: 4
    2: 5 3
    1: 4
    1: 3
    1: 4
    4: 6 8 1 5
    1: 4

    Sample Output:

    3
    4 3 1

    题目大意:

    有n个人,每个人喜欢k个活动,如果两个人有任意一个活动相同,就称为他们处于同一个社交网络。求这n个人一共形成了多少个社交网络~

    分析:

    并查集。先写好init、findFather、Union。

    1. 每个社交圈的结点号是人的编号,而不是课程。课程是用来判断是否处在一个社交圈的。

    2. course[t]表示任意一个喜欢t活动的人的编号。如果当前的课程t,之前并没有人喜欢过,那么就course[t] = i,i为它自己的编号,表示i为喜欢course[t]的一个人的编号

    3. course[t]是喜欢t活动的人的编号,那么findFather(course[t])就是喜欢这个活动的人所处的社交圈子的根结点,合并根结点和当前人的编号的结点i。即Union(i, findFather(course[t])),把它们处在同一个社交圈子里面

    4. isRoot[i]表示编号i的人是不是它自己社交圈子的根结点,如果等于0表示不是根结点,如果不等于0,每次标记isRoot[findFather(i)]++,那么isRoot保存的就是如果当前是根结点,那么这个社交圈里面的总人数

    5. isRoot中不为0的编号的个数cnt就是社交圈圈子的个数

    6. 把isRoot从大到小排列,输出前cnt个,就是社交圈人数的从大到小的输出顺序

    原文链接:https://blog.csdn.net/liuchuo/article/details/52191082

    题解

    image
    image

    #include <bits/stdc++.h>
    using namespace std;
    const int N=1010;
    int father[N]; //存放父亲结点
    int isRoot[N]; //记录每个结点是否作为某个集合的根节点
    int course[N];
    
    int findFather(int x){ //查找x所在集合的根节点
        int a=x;
        while(x!=father[x]){
            x=father[x];
        }
        //路径压缩
        while(a!=father[a]){
            int z=a;
            a=father[a];
            father[z]=x;
        }
        return x;
    }
    void Union(int a,int b){
        int faA=findFather(a);
        int faB=findFather(b);
        if(faA!=faB){
            father[faA]=faB;
        }
    }
    void init(int n){
        for(int i=1;i<=n;i++){
            father[i]=i;
            isRoot[i]=false;
        }
    }
    bool cmp(int a,int b){
        return a>b;
    }
    int main() {
        int n,k,h;
        cin>>n;
        init(n);
        for(int i=1;i<=n;i++){
            scanf("%d:",&k);
            for(int j=0;j<k;j++){
                cin>>h; //输入i号人喜欢的活动
                if(course[h]==0) //如果活动h第一次有人喜欢
                    course[h]=i; //令i喜欢活动h
                Union(i,findFather(course[h])); //合并
            }
        }
        for(int i=1;i<=n;i++)
            isRoot[findFather(i)]++; //i的根节点是findFather(i),人数+1
        int ans=0; //记录集合数目
        for(int i=1;i<=n;i++){
            if(isRoot[i]!=0)
                ans++; //只统计isRoot[i]不为0的
        }
        cout<<ans<<endl;
        sort(isRoot+1,isRoot+n+1,cmp);
        for(int i=1;i<=ans;i++){
            cout<<isRoot[i];
            if(i<ans) cout<<" ";
        }
    	return 0;
    }
    
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  • 原文地址:https://www.cnblogs.com/moonlight1999/p/15854090.html
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