POJ 2236 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

题意：发生了一次地震，网络系统瘫痪，那么现在就需要ACMer来修理网络了，有n个网络点，我们知道它们所在的位置，我们只能进行两次操作，一个是修理这个网络，一个是检查两个网络点是否可以联系，可以联系有两种情况：
1.两个网络点都被修理了，且两个点可以直接进行联系（两点距离小于d）；
2.两个网络点不能直接联系但是可以间接联系，前提是这些间接性的点都是被修理过的。

#include<stdio.h>

const int N=1010;

int f[N], vis[N], n; ///vis数组标记该网络点是否被修理过

void Init()
{
    int i;

    for (i = 0; i < n; i++)
    {
        vis[i] = 0; ///一开始网络点都是坏的
        f[i] = i;
    }
}

int Find(int x)
{
    if (f[x] != x)
        f[x] = Find(f[x]);

    return f[x];
}

int main ()
{
    int D, a[N], b[N], d, i, x, y, p, q;
    char s[10];

    scanf("%d%d", &n, &D);

    Init();

    for (i = 1; i <= n; i++)
        scanf("%d%d", &a[i], &b[i]);
    while (scanf("%s", s) != EOF)
    {
        if (s[0] == 'O')
        {
            scanf("%d", &x);
            vis[x] = 1;
            for (i = 1; i <= n; i++)
            {
                p = Find(x);
                q = Find(i);

                d = (a[x]-a[i])*(a[x]-a[i]) + (b[x]-b[i])*(b[x]-b[i]);

                if (d <= D*D && vis[i] && p != q) ///如果两个点都被修理过且距离小于D，那就把它们放在一个集合中，代表可以联系了
                    f[p] = q;
            }
        }
        else
        {
            scanf("%d%d", &x, &y);

            p = Find(x);
            q = Find(y);

            if (p == q && vis[x] && vis[y]) printf("SUCCESS
"); ///那么判断两个点是否可以联系只需要判断它们是否在一个集合就行了
            else printf("FAIL
");
        }
    }

    return 0;
}