POJ 2236 Wireless Network

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

题目大意： 给你n台坏掉的电脑，让你修理电脑，并且询问两台修好的电脑之间的距离是否小于d

#include <stdio.h>
#include <algorithm>
#include <iostream>
#define N 10000
using namespace std;
int n,d,fi[N],se[N],computer,computer2;
int f[N],repair[N];
void intit(int n)
{
    for(int i=0;i<=n;i++)
    {
        f[i]=i;
        repair[i]=0;
    }
}
int find_father(int x)
{
    if(x==f[x])
        return x;
    else return find_father(f[x]);
}
void unin(int x,int y)
{
    int xx,yy;
    xx= find_father(x);
    yy= find_father(y);
    if(xx!=yy)
        f[yy]=xx;
    //dis[yy][xx]++;
}
int judge(int cop1,int cop2)
{
    int disx=fi[cop1]-fi[cop2];
    int disy=se[cop1]-se[cop2];

    if((disx*disx+disy*disy) <=d*d)
        return 1;
    else
        return 0;
}
int main()
{

    scanf("%d %d",&n,&d);
    intit(n);
    for(int i=1;i<=n;i++)
    {
        scanf("%d %d",&fi[i],&se[i]);  //一开始输入的是每台电脑的坐标，亲们别搞错了。
    }
    char op;

    while(cin>>op)
    {
            if(op=='O')
            {
                scanf("%d",&computer);
                repair[computer]=1;
                for(int i=1;i<=n;i++)
                {
                    if(i!=computer&&repair[i]==1&&judge(i,computer))
                        unin(i,computer);  //把修好的电脑并且满足条件的电脑连接。
                }
            }
            else if(op=='S')
            {
                scanf("%d %d",&computer,&computer2);
                if(find_father(computer)==find_father(computer2))
                        printf("SUCCESS\n");
                    else printf("FAIL\n");
            }
    }
    return 0;
}


写的不好，请多提宝贵意见！！！