1107 Social Clusters (30 point(s))
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:
K
i
: h
i
[1] h
i
[2] … h
i
[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: 4
Sample Output:
3
4 3 1
#include<bits/stdc++.h>
using namespace std;
int father[1010];
int isRoot[1010];//记录每个社群有几个人
int cluster[1010]={0};//记录活动编号
int n,m,k;
bool cmp(int a,int b){
return a>b;
}
int findFather(int x){
int a=x;
while(x!=father[x]){
x=father[x];
}
while(a!=father[a]){//路径压缩
int z=a;
a=father[a];
father[z]=x;
}
return x;
}
void Union(int a,int b){
int faA=findFather(a);
int faB=findFather(b);
if(faA!=faB)
father[faA]=faB;
}
void init(int n){
for(int i=1;i<=n;++i){
father[i]=i;
isRoot[i]=0;
}
}
int main(){
scanf("%d",&n);
init(n);
for(int i=1;i<=n;++i){
scanf("%d:",&m);
for(int j=0;j<m;++j){
scanf("%d ",&k);
if(cluster[k]==0) cluster[k]=i;//若没有这项活动
Union(i,findFather(cluster[k]));
}
}
for(int i=1;i<=n;++i){
isRoot[findFather(i)]++;
}
int ans=0;
for(int i=1;i<=n;++i)
if(isRoot[i]) ans++;
sort(isRoot+1,isRoot+n+1,cmp);
printf("%d\n",ans);
for(int i=1;i<=ans;++i){
if(i!=1) printf(" ");
printf("%d",isRoot[i]);
}
}