Wireless Network （并查集）

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在东南亚地震后的背景下，一支医疗团队建立了临时无线网络，但随后的余震导致所有计算机损坏。随着计算机逐一修复，网络逐渐恢复正常。由于硬件限制，每台计算机只能与距离不超过特定范围内的其他计算机直接通信。本文探讨了如何通过修复操作和测试，确保计算机间能够成功通信。

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

Sample Output

FAIL
SUCCESS

答案：

#include <iostream>
#include <cstdio>
#include <cstring>
#include <cmath>
#define INF 0x3f3f3f3f
struct node
{
int x;
int y;
int data;
} p[1005],a,b;
double dis[1005][1005];
int f[1005];
int vis[1005];
double dist(struct node a,struct node b)
{
return sqrt((a.x-b.x)*1.0*(a.x-b.x)+(a.y-b.y)*1.0*(a.y-b.y));
}
int getf(int x)
{
if(x==f[x])
return x;
f[x] = getf(f[x]);
return f[x];
}
int join(struct node a,struct node b)
{
int xx = getf(a.data);
int yy = getf(b.data);
if(xx != yy)
f[xx] = yy;///因为做这个题发现这个顺序很重要！！！
}
int main()
{
int n,d;
memset(dis,INF,sizeof(dis));
memset(vis,0,sizeof(vis));
scanf("%d%d",&n,&d);
for(int i=1; i<=n; i++)
{
scanf("%d%d",&p[i].x,&p[i].y);
p[i].data = i;
f[i] = i;
}
char ch;
int a;
int b,c;
getchar();
while(~scanf("%c",&ch))
{
if(ch=='O')
{
scanf("%d",&a);
for(int j=1; j<=n; j++)
{
if(vis[j])
{
dis[a][j] = dist(p[a],p[j]);
if(dis[a][j]<=(double)d)
{
join(p[a],p[j]);
}
}
}
vis[a] = 1;
}
else if(ch=='S')
{
scanf("%d%d",&b,&c);
if(getf(b)==getf(c))
printf("SUCCESS\n");
else printf("FAIL\n");
}
}
return 0;
}