PAT1107:Social Clusters

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:

Sample Output:

3
4 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<iostream>
#include<vector>
#include<algorithm>
#include<iomanip>
using namespace std;
vector<int> source(1001,0);
vector<int> 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<int> 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];
       }

   }
}