POJ-2236 Wireless Network

Wireless Network

Time Limit: 10000MS		Memory Limit: 65536K
Total Submissions: 60216		Accepted: 24238

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

Source

POJ Monthly,HQM

#include<string.h>
#include<stdio.h>
#include<iostream>
#include<cmath>
using namespace std;
int pre[31005];	
int n;
double d;
struct node{
	double x;
	double y;
}p[1005];
bool dis(node a,node b){//返回a，b两点之间距离小于d的 
	return sqrt((a.x -b.x )*(a.x -b.x )+(a.y -b.y )*(a.y -b.y ))<=d;
}
int find(int x){
	if(pre[x]==x)
	    return pre[x];
	else
	    return pre[x]=find(pre[x]);
}
void merge(int x,int y){
	int fx=find(x);
	int fy=find(y);
	if(fx!=fy){
		pre[fx]=fy;
	}
}
int main(){
	scanf("%d%lf",&n,&d);//输入n和d 
	for(int i=1;i<=n;i++){//输入下面的n个点 
		scanf("%lf%lf",&p[i].x ,&p[i].y );
	}
	char c;
	while(~scanf("%c",&c)){
		if(c=='O'){//o修复 
			int x;
			scanf("%d",&x);
			pre[x]=x;
			for(int i=1;i<=n;i++){
			    if(pre[i]&&dis(p[i],p[x])){//看看能由这个修复点再链接哪些点 
			    	merge(i,x);
				}				
			}	
		}
		if(c=='S'){///s询问 
			int x,y;
			scanf("%d%d",&x,&y);
			if(find(x)==find(y))
			    printf("SUCCESS\n");
			else
			    printf("FAIL\n");  
		}
	}
	return 0;
}