Friends(采用树结构的非线性表编程)

最新推荐文章于 2025-06-14 14:30:07 发布

转载最新推荐文章于 2025-06-14 14:30:07 发布 · 99 阅读

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原文链接：http://www.cnblogs.com/lengxia/p/4387820.html

本文介绍了一个算法问题，即计算在一个城镇中最大的朋友圈规模。通过输入城镇公民数量及已知的朋友关系，利用并查集的方法来找出最大的朋友圈包含多少成员。

Description

There is a town with N citizens. It is known that some pairs of people are friends. According to the famous saying that “The friends of my friends are my friends, too” it follows that if A and B are friends and B and C are friends then A and C are friends, too.

Your task is to count how many people there are in the largest group of friends.

Input

Input consists of several datasets. The first line of the input consists of a line with the number of test cases to follow. The first line of each dataset contains tho numbers N and M, where N is the number of town's citizens (1≤N≤30000) and M is the number of pairs of people (0≤M≤500000), which are known to be friends. Each of the following M lines consists of two integers A and B (1≤A≤N, 1≤B≤N, A≠B) which describe that A and B are friends. There could be repetitions among the given pairs.

Output

The output for each test case should contain one number denoting how many people there are in the largest group of friends.

Sample Input

Sample Output

3 2

1 2

2 3

10 12

1 2

3 1

3 4

5 4

3 5

4 6

5 2

2 1

7 10

1 2

9 10

8 9

#include <iostream>
#include <algorithm>
using namespace std;
int p[30000],c[30000];
int find(int x)
{return p[x]==x ? x : p[x]=find(p[x]);
}
int main()
{
int t,i,x,y;
cin>>t;
while(t--)
{
  int m,n;
  cin>>m>>n;
  for(i=1;i<=m;i++)
  {p[i]=i;c[i]=0;}
  for(i=1;i<=n;i++)
  {
cin>>x>>y;
  x=find(x);
  y=find(y);
  if(x!=y)
   p[x]=y;
  }
  for(i=1;i<=m;i++)
  {x=find(i);
  c[x]++;
  }

  int max=0;
        for(int i=1; i<=m; i++)
            if(c[i]>max)
                max=c[i];


   cout<<max<<endl;

}
return 0;
}