1107 Social Clusters （30 分）

最新推荐文章于 2020-10-21 09:52:03 发布

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本文介绍了一种使用并查集算法解决社交网络中寻找共同兴趣群体的问题。通过输入每个人的爱好列表，算法能够找出所有具有共同爱好的人群集群，并统计每个集群的人数。

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:

Sample Output:

3
4 3 1

思路：首先根据题意，输入数据n表示n个人，每行给出该人喜欢的活动，求有多少不同的集合以及每个集合的人数，因此可以用并查集去做，设置ans[h] 记录任意喜欢该活动的人的编号，findfather(ans[h])就是这个人所在网络的节点。故只需要合并当前该人编号和他喜欢活动的人（ans[h]）所在在的根节点即可。

集合计数统计有多少根就可以，isroot[] ！=0 则为根。排序isroot[] 找出每个集合人数。

#include<bits/stdc++.h>
using namespace std;
const int maxn=1100;
int father[maxn];
int isroot[maxn]={0};
int ans[maxn]={0},n;
bool cmp(int a, int b)
{
    return a>b;
}
int findfather(int x)
{
    while(x!=father[x]) x=father[x];
    return x;
}
void unio(int a,int b)
{

    int fa=findfather(a);
    int fb=findfather(b);
    if(fa!=fb){father[fa]=fb;}
}
void init(int n){
    for(int i=1;i<=n;i++)
    {
        father[i]=i;
        isroot[i]=false;
    }
}
int main()
{

    scanf("%d",&n);
    init(n);
    int tmp,h;
    for(int i=1;i<=n;i++)
    {
        scanf("%d:",&tmp);
        for(int j=0;j<tmp;j++)
        {
           scanf("%d",&h);
           if(ans[h]==0) ans[h]=i;
           unio(i,findfather(ans[h]));
        }
    }
    for(int i=1;i<=n;i++)
    {
        isroot[findfather(i)]++;
    }
    int cnt=0;
    for(int i=1;i<=n;i++)
    {
        if(isroot[i]!=0) cnt++;
    }
    printf("%d\n",cnt);
    sort(isroot+1,isroot+n+1,cmp);
    for(int i=1;i<=cnt;i++)
    {
        if(i<cnt) printf("%d ",isroot[i]);
        else printf("%d",isroot[i]);
    }
    return 0;
}