poj 2236 Wireless NetWork

Wireless Network

Description

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: 
1. "O p" (1 <= p <= N), which means repairing computer p. 
2. "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

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題意 ：給一些電腦的座標,有兩種操作 'O p'即修復p點 'S p q'詢問p,q的是否連通；

      連通的條件是：兩點之間的距離小於d,或者可以通過中間節點連接；並且兩點都是修復過的；

思路：判斷當前修復好的節點(是一個只有一個節點的連通分量)，和之前修復好的節點(要遍歷修復好的所有的節點)之間的距離D ,如果D<=d,那麼就將此分量和之前節點所在的連通分量並在一起；

#include<stdio.h>
#include<string.h>
#include<math.h>
struct node
{
	int x,y;
	int pre;
}node[1010];


double dou(int x)
{
	return (x)*(x)*1.0; 
}
double distance(int x,int y,int k)
{
	return sqrt(dou(node[k].x - x) + dou(node[k].y - y));
}


int find(int k)
{
	while(k!=node[k].pre)
		k = node[k].pre;
	return k;
}
int main()
{
	int n,d;
	int i,j,k,x;
	int a[1010];//數組a用於存放所有已修復好的節點；
	char ch;
	double tmp=0;
	while(scanf("%d%d",&n,&d)!=EOF)
	{
		memset(node,0,sizeof(node));
		for(i=1;i<=n;i++)
		{
			scanf("%d%d",&node[i].x,&node[i].y);
			node[i].pre = 0;
		}
		memset(a,0,sizeof(a));
		x = 1;
		getchar();
		while(scanf("%c",&ch)!=EOF)
		{
			if(ch=='O')
			{
				scanf("%d%*c",&k);
				node[k].pre = k;//表示此點被修復好了
				for(j=1;j<x;j++)//和之前修復好的節點進行判斷
				{
					tmp = distance(node[k].x,node[k].y,a[j]);
					if(tmp<=d*1.0)
					{
						int fa1,fa2;
						fa1 = find(a[j]);
						fa2 = find(k);
						if(fa1 != fa2)
							node[fa1].pre = fa2;
					}
				}
				a[x++] = k;
			}
			else if(ch=='S')
			{
				int p1,p2;
				scanf("%d%d%*c",&p1,&p2);
				if(find(p1)==find(p2))
					printf("SUCCESS\n");
				else
					printf("FAIL\n");
			}
		}
	}
	return 0;
}