并查集复习进阶版

An earthquake takes place in Southeast Asia. The ACM (Asia Cooperated Medical team) have set up a wireless network with the lap computers, but an unexpected aftershock attacked, all computers in the network were all broken. The computers are repaired one by one, and the network gradually began to work again. Because of the hardware restricts, each computer can only directly communicate with the computers that are not farther than d meters from it. But every computer can be regarded as the intermediary of the communication between two other computers, that is to say computer A and computer B can communicate if computer A and computer B can communicate directly or there is a computer C that can communicate with both A and B.

In the process of repairing the network, workers can take two kinds of operations at every moment, repairing a computer, or testing if two computers can communicate. Your job is to answer all the testing operations.
Input
The first line contains two integers N and d (1 <= N <= 1001, 0 <= d <= 20000). Here N is the number of computers, which are numbered from 1 to N, and D is the maximum distance two computers can communicate directly. In the next N lines, each contains two integers xi, yi (0 <= xi, yi <= 10000), which is the coordinate of N computers. From the (N+1)-th line to the end of input, there are operations, which are carried out one by one. Each line contains an operation in one of following two formats:

“O p” (1 <= p <= N), which means repairing computer p.
“S p q” (1 <= p, q <= N), which means testing whether computer p and q can communicate.

The input will not exceed 300000 lines.
Output
For each Testing operation, print “SUCCESS” if the two computers can communicate, or “FAIL” if not.
Sample Input
4 1
0 1
0 2
0 3
0 4
O 1
O 2
O 4
S 1 4
O 3
S 1 4
Sample Output
FAIL
SUCCESS
ac代码：

#include<stdio.h>
#include<math.h>
int n,d;
int x[2000];
int par[2000];
void ini()
{
   
	int i;
	for(i=1;i<=n;i++)
	par[i]=i;
}
int y[2000];
char a[2];
int repair[2005];
int getf(int n)
{
   
	if(n==par[n])
	return n;
	else
	par[n]=getf(par[n]);
	return par[n];
}
void union1(int x,int y)
{
   
	int t1,t2;
	t1=getf(x);
	t2=getf(y);
	if(t1!=t2)
	par[t2]=t1;
}
double ju(int a,int b)
{
   
	return sqrt(1.0*(x[a]-x[b])*(x[a]-x[b])+(y[a]-y[b])*(y[a]-y[b]));
}
int main()
{
   

	int i;
	scanf("%d%d",&n,&d);
	for(i=1;i<=n;i++)//i从1开始注意细节
	scanf("%d%d",&x[i],&y[i]);
	int len=0;
	int p;	int q;
	ini();//别忘了初始化数组
	while(scanf("%s",a)!=EOF)//scanf刚好不吃空格，QAQ.
	{
   
		switch(a[0]){
   
			case 'O':	scanf("%d",&p);//输入编号为p的计算机
			repair[len++]=p;	//len++
			for(i=0;i<=len-1;i++)
			{
   
			if(ju(p,repair[i])<=double(d)&&repair[i]!=p)//距离要小于d并且不要等于本身,注意精度。
				union1(repair[i],p);//将他们俩合并一下
			}
			break;
			case 'S': scanf("%d%d",&p,&q);
				if(getf(p)==getf(q))
				printf("SUCCESS\n");
				else
				printf("FAIL\n");
				break;
			default: break;
					}
	}
		return 0;
 }