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:

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:

83: 2 7 101: 42: 5 31: 41: 31: 44: 6 8 1 51: 4

Sample Output:

34 3 1

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将有相同的兴趣（一个或多个相同）的人归类成一个集体，输出集体的个数和各个集体的人数。这里用了两个数组来储存信息，一个储存各个人有哪些兴趣，另一个储存各个兴趣对应的哪些人。然后用bfs，先从某一个兴趣出发，找出对应的人（期间要更新当前集合的人数），然后再根据这些人得到相关的兴趣，然后再从兴趣得到人，从人得到兴趣......这样循环直到当前的兴趣队伍空了。为了防止无限循环，要用两个数组保存哪些人和哪些兴趣被访问过。

代码：

#include <iostream>#include <cstring>#include <cstdio>#include <cstdlib>#include <vector>#include <set>#include <map>#include <queue>#include <algorithm>using namespace std;int main(){int n;cin>>n;vector<vector<int> >hob(1001);vector<vector<int> >peo(n);int Max=0;for(int i=0;i<n;i++){int k;cin>>k;getchar();for(int j=0;j<k;j++){int h;cin>>h;Max=max(Max,h);peo[i].push_back(h);hob[h].push_back(i);}}vector<bool>isvis_h(Max+1,false);vector<bool>isvis_p(n,false);vector<int>res;for(int i=1;i<=Max;i++){if(hob[i].size()==0||isvis_h[i]) continue;int count=0;queue<int>que;que.push(i);while(!que.empty()){int tmp=que.front();que.pop();isvis_h[tmp]=true;for(int j=0;j<hob[tmp].size();j++){if(!isvis_p[hob[tmp][j]]){count++;isvis_p[hob[tmp][j]]=true;for(int k=0;k<peo[hob[tmp][j]].size();k++){if(!isvis_h[peo[hob[tmp][j]][k]]){que.push(peo[hob[tmp][j]][k]);}}}}}res.push_back(count);}cout<<res.size()<<endl;sort(res.begin(),res.end(),greater<int>());cout<<res[0];for(int i=1;i<res.size();i++){cout<<" "<<res[i];}}