POJ 2236 Wireless Network 并查集

Wireless Network 并查集

Crawling in process... Crawling failed Time Limit:10000MS     Memory Limit:65536KB     64bit IO Format:%I64d & %I64u

Submit Status Practice POJ 2236

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
说说题目大意，这题是先给出n和d表示计算机的台数和两台计算机能够连上的最大距离，然后给出一些操作，O表示修编号电脑，S表示询问编号为k，m的电脑是否能够联通，联通的条件是两台电脑都修好了而且两台电脑的距离不超过最大距离。。
上个渣渣的并查集代码
#include<stdio.h>
#include<string.h>
#include<iostream>
#include<algorithm>
using namespace std;
int n;
struct point
{
    int x,y;
}qq[1010];
int pre[10000],p[10000];
int find(int x)
{
    if(x!=pre[x])
    pre[x]=find(pre[x]);//递归求父亲节点
    return pre[x];
}
void get(int x,int y)
{
    int xx=find(x);
    int yy=find(y);
    pre[yy]=pre[xx];、、路径压缩
}
void init()
{
    for(int i=1;i<=n;i++)
    {
        pre[i]=i;
        
    }
}
int dis(int i,int j)
{
    return (qq[i].x-qq[j].x)*(qq[i].x-qq[j].x)+(qq[i].y-qq[j].y)*(qq[i].y-qq[j].y);
}
int main()
{
    int d;
    cin>>n>>d;
    for(int i=1;i<=n;i++)
    {
        cin>>qq[i].x>>qq[i].y; 
    }
    init();
    char s[5];
    int op,k,m;
    int count=0;
    while(scanf("%s",s)!=EOF)//最好用这个字符串输入，自动忽略空格
    {
        if(s[0]=='O') 
        {
            cin>>k;
            for(int i=0;i<count;i++)
            {
                if(dis(p[i],k)<=d*d)
                get(p[i],k);
            }
                p[count++]=k;
        }
        else
        {
            cin>>k>>m; 
            if(find(k)!=find(m))
            printf("FAIL\n");
            else
            printf("SUCCESS\n");
        }
    }
    return 0;
}