• 1107 Social Clusters (30 分)


    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 (≤), 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​​ (>) is the number of hobbies, and [ 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

    思路:

    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。

     
    #include <bits/stdc++.h>
    using namespace std;
    int p[10010],a[10010];
    int n,k,m,h;
    char c;
    bool cmp(int a,int b)
    {
        return a>b;
    }
    
    int found(int x)
    {
        if(p[x] == x)
            return x;
        return found(p[x]);
    }//found函数返回的是
    void unite(int a,int b)
    {
        int x = found(a);
        int y = found(b);
        if(x!=y)
        {
            p[y] = x;
        }
    }//unite函数
    int main()
    {
        scanf("%d",&n);
        for(int i=0;i<=n;i++)
            p[i] = i;
        for(int i=1;i<=n;i++)
        {
            scanf("%d %c",&k,&c);
            for(int j=1;j<=k;j++)
            {
                scanf("%d",&h);
                if(a[h] == 0)
                {
                    a[h] = i;
                }
                else
                {
                    unite(found(a[h]),i);
                }
            }
        }
        int cnt = 0;
        int count1[1010] = {0};
        for(int i=1;i<=n;i++)
        {
            count1[found(i)]++;
        }
        for(int i=1;i<=n;i++)
        {
            if(count1[i]!=0)
                cnt++;
        }
        cout<<cnt<<endl;
        sort(count1+1,count1+n+1,cmp);
        for(int i=1;i<=cnt;i++)
        {
            if(i==1)
                cout<<count1[i];
            else
                cout<<" "<<count1[i];
        }
        return 0;
    }
     
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  • 原文地址:https://www.cnblogs.com/tonyyy/p/10544331.html
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