【POJ】2236 - Wireless Network（并查集）

Wireless Network Time Limit:10000MS Memory Limit:65536KB 64bit IO Format:%I64d & %I64u

Submit Status

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

做这道题，最担心的就是TLE，因为每次进行 O 操作，都需要遍历一遍，如果电脑多， O 操作多，应该是非常费时间的吧，

还好顺利AC，毕竟限制时间是10s，我用了3.032s

代码如下：

#include <cstdio>
#include <cstring>
#include <cmath>
#include <algorithm>
using namespace std;
int f[1111];
struct node
{
	double x,y;
	bool flag;
}dot[1111];
double dis(node a,node b)
{
	return sqrt((a.x-b.x)*(a.x-b.x)+(a.y-b.y)*(a.y-b.y));
}
int find (int x)
{
	if (x!=f[x])
		f[x]=find(f[x]);
	return f[x];
}
void join(int x,int y) 
{
	int fx,fy;
	fx=find(x);
	fy=find(y);
	if (fx!=fy)
		f[fx]=fy;
}
int main()
{
	int n;
	double d;
	char op;		//操作 
	scanf ("%d %lf",&n,&d);
	for (int i=1;i<=n;i++)
		f[i]=i;
	for (int i=1;i<=n;i++)
		scanf ("%lf %lf",&dot[i].x,&dot[i].y);
	getchar();
	while (~scanf ("%c",&op))
	{
		if (op=='O')
		{
			int t;
			scanf ("%d",&t);
			getchar();
			dot[t].flag=true;
			for (int i=1;i<=n;i++)
			{
				if (dis(dot[i],dot[t])<=d && dot[i].flag && dot[t].flag)
					join(t,i);
			}
		}
		else
		{
			int t1,t2;
			scanf ("%d %d",&t1,&t2);
			getchar();
			if (find(t1)==find(t2))
				printf ("SUCCESS\n");
			else
				printf ("FAIL\n");
		}
	}
	return 0;
}