【POJ 2236 --- Wireless Network】并查集

题目来源：点击进入【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

解题思路

每次修理好一台计算机的时候就遍历一下所有修好的计算机，看距离是否<=d，如果符合说明可以连通，将两台计算机所在集合合并。
每次检查的时候判断一下这两台计算机是否在同一集合中即可。

AC代码：

#include <iostream>
#include <algorithm>
using namespace std;
#define SIS std::ios::sync_with_stdio(false),cin.tie(0),cout.tie(0)
#define endl '\n'
const int MAXN = 1e3+5;
int pre[MAXN],arr[MAXN];

struct Node
{
    int x,y;
}node[MAXN];

void init()
{
    for(int i=0;i<MAXN;i++)
        pre[i]=i;
}

int _find(int x)
{
    if(x==pre[x]) return x;
    return pre[x]=_find(pre[x]);
}

void unite(int x,int y)
{
    x=_find(x);
    y=_find(y);
    if(x!=y) pre[x]=y;
}

int dis(Node &x,Node &y)
{
    return (x.x-y.x)*(x.x-y.x)+(x.y-y.y)*(x.y-y.y);
}

int main()
{
    SIS;
    int n,d;
    init();
    cin >> n >> d;
    for(int i=1;i<=n;i++)
        cin >> node[i].x >> node[i].y;
    char ch;
    int num=0,x,y;
    while(cin >> ch)
    {
        if(ch=='O')
        {
            cin >> x;
            arr[num++]=x;
            for(int i=0;i<num-1;i++)
                if(dis(node[arr[i]],node[x])<=d*d)
                    unite(arr[i],x);
        }
        else
        {
            cin >> x >> y;
            if(_find(x)==_find(y)) cout << "SUCCESS" << endl;
            else cout << "FAIL" << endl;
        }
    }
    return 0;
}