PAT 1107. Social Clusters (30)

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

并查集

每个人都有很多很多的爱好，如果两个人有相同的爱好，他们就属于统统一个集合。

求集合数和每个集合的人数，降序输出。

将相同爱好的人加入并查集，统计祖先的个数就可以了。

#include<bits/stdc++.h>

using namespace std;
const int N=1123;
int f
,ans
;
vector<int>v
;
map<int,int>mp;

bool cmp(int a,int b)
{
return a>b;
}

int Find(int x)
{
if(x!=f[x])f[x]=Find(f[x]);
return f[x];
}

void Join(int x,int y)
{
int fx=Find(x);
int fy=Find(y);
if(fx!=fy)
{
f[fx]=fy;
}
}

int main()
{
int n;
scanf("%d",&n);
for(int i=0; i<=1001; i++)
{
f[i]=i;
}
for(int i=1; i<=n; i++)
{
int k;
scanf("%d:",&k);
for(int j=0; j<=k-1; j++)
{
int x;
scanf("%d",&x);
v[x].push_back(i);
}
}
for(int i=1; i<=1001; i++)
{
if(v[i].size()>1)
{
for(int j=1; j<=(int)v[i].size()-1; j++)
{
Join(v[i][0],v[i][j]);
}
}
}
for(int i=1; i<=n; i++)
{
int fx=Find(i);
if(mp.find(fx)!=mp.end())mp[fx]++;
else mp[fx]=1;
}
map<int,int>::iterator it;
int cnt=0;
for(it=mp.begin(); it!=mp.end(); it++)
{
ans[cnt++]=it->second;
}
sort(ans,ans+cnt,cmp);
printf("%d\n",cnt);
for(int i=0; i<=cnt-1; i++)
{
if(i==cnt-1)printf("%d\n",ans[i]);
else printf("%d ",ans[i]);
}
return 0;
}