pat-a 1107. Social Clusters (30)

并查集，我写的多了一点。

简单说下就行，两个人如果有共同的爱好就是一个圈子里面的。开始错误原因在于压缩路径不够彻底。因为最后判断有几个圈子时没有去寻找根直接判断父节点，因此要压缩到底才能直接判断。

#include<cstdio>
#include<algorithm>
#include<functional>
using namespace std;
int num[1010][1010];
int p[1010];
int visit[1010];
void init(){
for(int i=0;i<1010;++i) p[i]=i;
}
int find(int x){
if(p[x]==x) return x;
else return p[x]=find(p[x]);
}
void merge(int i,int j){
int x=find(i);
int y=find(j);
if(x!=y){
if(x>y) p[x]=y;
else p[y]=x;
}
}
int main(){
int n,t,a,ans=0,flag=0,k=0;
int c[1010];
scanf("%d",&n);
init();
for(int i=0;i<n;++i){
scanf("%d:",&t);
while(t--){
scanf("%d",&a);
num[i][a]=1;
for(int j=0;j<i;++j){
if(num[j][a]){
merge(i,j);
}
}
}
}
for(int i=0;i<n;++i){
for(int j=0;j<i;++j){
if(find(i)==find(j)) merge(i,j);
}
}
//for(int i=0;i<n;++i) printf("%d ",p[i]);
for(int i=0;i<n;++i){
visit[p[i]]++;
if(visit[p[i]]==1) ans++;
}
printf("%d\n",ans);
for(int i=0;i<n;++i){
if(visit[i]){
c[k++]=visit[i];
}
}
sort(c,c+k,greater<int>());
for(int i=0;i<k;++i){
if(flag)printf(" %d",c[i]);
else{
flag=1;
printf("%d",c[i]);
}
}
return 0;
}

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