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

思路   并查集问题，course数组保存最先选择该课程的，source保存每个人关联的源点，然后查找合并就行。 代码

#include
     
      #include
      
       #include
       
        #include
        
         using namespace std;vector
         
           source(1001,0);vector
          
            course(1001,0);int findSource(int a){ int t = a; while(source[a] != a) { a = source[a]; } while(source[t] != t) { int tmp = t; t = source[t]; source[tmp] = a; } return a;}void Union(int a,int b){ a = findSource(a); b = findSource(b); if( a != b) source[a] = b;}bool cmp(int a,int b){ return a > b;}int main(){ int N; while(cin >> N) { vector
           
             root(N + 1,0); for(int i = 1;i <= N;i++) { source[i] = i; } for(int i = 1;i <= N;i++) { int k; scanf("%d : ",&k); for(int j = 1;j <= k;j++) { int h; cin >> h; if(course[h] == 0) course[h] = i; Union(i,findSource(course[h])); } } for(int i = 1;i <= N;i++) { ++root[findSource(i)]; } int cnt = 0; for(int i = 1;i <= N;i++) { if(root[i] > 0) cnt++; } cout << cnt << endl; int flag = 0; sort(root.begin(),root.end(),cmp); for(int i = 0;i < cnt;i++) { if(flag++ != 0) cout << " "; cout << root[i]; } }}