UVA 10608 Friends

Friends（8.4.1）

Crawling in process...Crawling failedTime Limit:3000MS    Memory Limit:0KB     64bit IO Format:%lld & %llu

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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
2 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	3 6

 

#include <iostream>
#include <cstring>
#define maxn 30005
using namespace std;
int f[maxn],ans[maxn];
int find(int x)
{
    if(f[x]==x)
        return x;
    else
        return(f[x]=find(f[x]));	 
}
int main()
{
    int N;
	cin>>N;
    while(N--)
    { 
		int n,m;
		cin>>n>>m;
		
		int i;
		for(i=1;i<=n;i++)
			f[i]=i;
		
		memset(ans,0,sizeof(ans));
		
		for(i=1;i<=m;i++)
		{
			int x,y;
			cin>>x>>y;
			int fx,fy;
			fx=find(x);
			fy=find(y);
			if(fx!=fy)
				f[fx]=fy;
		}
		
		for(i=1;i<=n;i++)
			ans[find(i)]++;
		
		int max=0;
		for(i=1;i<=n;i++)
			if(ans[i]>max)
				max=ans[i];
			
			cout<<max<<endl;
    }
    return 0;
}

代码还没有ac。不管是在UVA上还是虚拟oj上都显示，In judge queue

表示非常无语，等了好长时间，虚拟oj显示  Judging Error 2。淡淡的忧桑。

。

更新：已经ac

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