PAT 1107. Social Clusters (30) 并查集

1107. Social Clusters (30)

时间限制1000 ms内存限制65536 kB代码长度限制16000 B判题程序Standard作者CHEN, YueWhen 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 aperson 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 spaceat 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

就是谁第一次访问到某习惯时，留下标记。后来访问到这个习惯的都是他儿子。

#include <iostream>#include<stdio.h>#include<string>#include<queue>#include<algorithm>#include<string.h>#include<vector>#include<map>#include<math.h>using namespace std;int hobby[1005]={0};int father[1005]={0};int cnt[1005]={0};bool cmp(int x,int y){return x>y;}vector<int> node[1005];int findpapa(int x){if(father[x]==x) return x;else{int zz=findpapa(father[x]);father[x]=zz;return zz;}}int main(){int n;cin>>n;for(int i=1;i<=n;i++){int tmp;cin>>tmp;getchar();for(int j=0;j<tmp;j++){int zz;cin>>zz;node[i].push_back(zz);}}for(int i=0;i<1004;i++)father[i]=i;for(int i=1;i<=n;i++){for(int j=0;j<node[i].size();j++){int tmp=node[i][j];if(hobby[tmp]==0) hobby[tmp]=father[i];else{father[father[i]]=hobby[tmp];findpapa(i);}}}for(int i=1;i<=n;i++)findpapa(i);for(int i=1;i<=n;i++)cnt[father[i]]++;sort(cnt+1,cnt+1+n,cmp);int p=1;while(cnt[p]!=0) p++;p--;cout<<p<<endl;int flag=0;for(int i=1;i<=p;i++){if(flag==1) cout<<' ';flag=1;cout<<cnt[i];}return 0;}