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 (≤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[*K**i*]
where *K**i* (>0) is the number of hobbies, and *h**i[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: 4Sample 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。
第一个喜欢这个活动的称为教主,之后喜欢这个活动的人和教主unite即可.
1.并查集写错了,返回一个等号写成了两个等号.
2.找每组属于谁的时候可能还没有压缩路径,所以应该统计find(i)而不是par[i],,找分为几组的时候才可以用par[i]==i
#include
#include
#define each(a,b,c) for(int a=b;a<=c;a++)
#define de(x) cout<<#x<<" "<<(x)<MS;
void init(int n)
{
for(int i=1;i<=n;i++)///初始化小错误
{
par[i]=i;
Rank[i]=0;
}
}
int find(int x)
{
return par[x]==x?x:par[x]=find(par[x]);///并查集的父节点满足par[x]==x,如果不是的话也要把par[x]更新,返回的是find(par[x])同时进行了路径的压缩
}
priority_queueQ;
void unite(int x,int y)
{
//de(x);
//de(y);
x=find(x);
y=find(y);
if(x==y)return;
if(Rank[x]b;}
int main()
{
int n;
scanf("%d",&n);
init(n);
cur=0;
each(i,1,n)
{
int k;
scanf("%d: ",&k);
while(k--)
{
int temp;
scanf("%d",&temp);
if(jiaozhu[temp]==0)
{
jiaozhu[temp]=i;
}
else
{
unite(jiaozhu[temp],i);
}
}
}
each(i,1,n)
{
root_cnt[find(i)]++;
}
for(int i=1;i<=n;i++)
{
if(root_cnt[i]!=0)
{
ans[cur++]=root_cnt[i];
}
}
sort(ans,ans+cur,cmp);
printf("%d\n",cur);
each(i,0,cur-1)
{
printf("%d",ans[i]);
printf(i==cur-1?"\n":" ");
}
}