UVa 10301

题目：有一些粘有胶水的圆环，散落在地上，问黏在一起最多的整体有几个环组成。

分析：并查集、计算几何。判断每两个圆环是否黏在一起即可，利用并查集记录。

注意：1.输出格式 0 rings。2.是圆环不是圆。3.并查集合并前要判断；把相同集合的合并会重复计数╮(╯▽╰)╭。

#include <iostream>
#include <cstdlib>
#include <cstdio>
#include <cmath>

using namespace std;

typedef struct rnode
{
	double x,y,r;
}ring;
ring R[105];

double dist( ring a, ring b )
{
	return sqrt((a.x-b.x)*(a.x-b.x)+(a.y-b.y)*(a.y-b.y));
}

//union_set
int  sets[105];
int  sums[105];
int  rank[105];

//返回根节点 
int  Find( int a )
{
	if ( a != sets[a] )
		sets[a] = Find( sets[a] );
	return sets[a];
}

//合并集合 
void Union( int a, int b )
{
	if ( rank[a] < rank[b] ) {
		sets[a]  = b;
		sums[b] += sums[a];
	}else {
		if ( rank[a] == rank[b] )
			rank[a] ++;
		sets[b]  = a;
		sums[a] += sums[b];
	}
}
//union_set end

int main()
{
	int n;
	while ( ~scanf("%d",&n) && n >= 0 ) {
		for ( int i = 0 ; i < n ; ++ i ) {
			scanf("%lf%lf%lf",&R[i].x,&R[i].y,&R[i].r);
			rank[i] = 1;sets[i] = i;sums[i] = 1;
		}
		
		for ( int i =  0  ; i < n ; ++ i )
		for ( int j = i+1 ; j < n ; ++ j ) {
			if ( dist( R[i], R[j] )-1e-6 < fabs( R[i].r+R[j].r ) &&
				 dist( R[i], R[j] )+1e-6 > fabs( R[i].r-R[j].r ) )
			if ( Find(i) != Find(j) )
				Union( Find( i ), Find( j ) );
		}
		
		int max = 0;
		for ( int i = 0 ; i < n ; ++ i )
			if ( max < sums[i] )
				max = sums[i];
		
		//万恶的输出啊，╮(╯▽╰)╭ 
		if ( max > 1 || !max )
			printf("The largest component contains %d rings.
",max);
		else printf("The largest component contains 1 ring.
");
	}
	return 0;
}