POJ2236 Wireless Network 并查集

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本文介绍了一个关于地震后无线网络重建的算法问题。该问题聚焦于如何通过修复不同位置的计算机来逐步恢复网络连接，并判断任意两台计算机是否可以通过直接或间接的方式进行通信。通过并查集的数据结构实现了对网络状态的有效管理和查询。

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Wireless Network

Time Limit: 10000MS		Memory Limit: 65536K
Total Submissions: 33412		Accepted: 13882

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

Sample Output

FAIL
SUCCESS

开一个数组记录修复状况，每有一个新的电脑修复就检测其D的距离内是否有好的电脑，将其连上机即可。

#include<stdio.h>
#include<string.h>
#define N 1010
struct node
{
   int x;
   int y;
}dis[N];
int f[N],flag[N];
int find(int x)
{
    if (x==f[x]) return x;
     else {f[x]=find(f[x]);return f[x];}
}
int distance(int a,int b)
{
    return ((dis[a].x-dis[b].x)*(dis[a].x-dis[b].x)+(dis[a].y-dis[b].y)*(dis[a].y-dis[b].y));
}
int main()
{
    int i,n,d;
    char s[20];
    scanf("%d%d",&n,&d);
    for (i=1;i<=n;i++)
    {
        scanf("%d%d",&dis[i].x,&dis[i].y);
        f[i]=i;flag[i]=0;
    }
    while (scanf("%s",s)!=EOF)
    {
        int a,b;
        if (s[0]=='O')
        {
            scanf("%d",&a);
            if (!flag[a])
            {
                flag[a]=1;
                for (i=1;i<=n;i++)
                {
                    if (flag[i]&&(distance(i,a)<=d*d)&&find(a)!=find(i)) f[f[a]]=f[i];
                }
            }
        }
        if (s[0]=='S')
        {
            scanf("%d%d",&a,&b);
            if (find(a)==find(b)) printf("SUCCESS\n"); else printf("FAIL\n");
        }
    }
}