【PAT】【Advanced Level】1107. Social Clusters (30)

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, Yue

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

原题链接：
https://www.patest.cn/contests/pat-a-practise/1107

思路：

先转化为图，再求连通分量的个数、规模

CODE：

#include<iostream>#include<vector>#include<map>#include<cstring>#include<cstdio>#include<algorithm>#define N 1010#define NN 1000100using namespace std;map<int,int> ind;vector<int> hb;vector<int> hbp[NN];vector<int> fri[N];bool fl[N];vector<int> res;int dfs(int n){int num=1;for (int i=0;i<fri[n].size();i++){if (fl[fri[n][i]]==1) continue;fl[fri[n][i]]=1;num+=dfs(fri[n][i]);}return num;}bool cmp(int a,int b){return a>b;}int main(){int n;scanf("%d",&n);for (int i=1;i<=n;i++){int m;scanf("%d:",&m);for (int j=0;j<m;j++){int t;scanf("%d",&t);if (ind[t]==0){hb.push_back(t);ind[t]=hb.size();hbp[ind[t]].push_back(i);}else{int si=hbp[ind[t]].size();for(int k=0;k<si;k++){//cout<<k<<endl;fri[i].push_back(hbp[ind[t]][k]);fri[hbp[ind[t]][k]].push_back(i);}hbp[ind[t]].push_back(i);}}}memset(fl,0,sizeof(fl));for (int i=1;i<=n;i++){if (fl[i]==1) continue;fl[i]=1;res.push_back(dfs(i));}sort(res.begin(),res.end(),cmp);printf("%d\n",res.size());for (int i=0;i<res.size();i++){if (i!=0) printf(" ");printf("%d",res[i]);}return 0;}