POJ - 2236 Wireless Network（并查集）

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：给N台未维修电脑，可连在一起的最大距离是d . 刚 开始所有电脑都是孤立的，输入N个电脑的坐标（x,y）; 输入多组命令，形如 O A :表示维修A电脑，A电脑就可以与在d距离范围内的其他电脑联机（并查集）; S A B ：问A，B是否在一个集合里

自己注意输入EOF

#include<iostream>
#include<stdio.h>
#define ll long long
#include<algorithm>
#include<string.h>
using namespace std;
const int maxn = 1001+10;
struct node
{
    int x,y;
}a[maxn];

int vis[maxn];//是否修了
int per[maxn];//并查集集合数组
int n,d,A,B,dis;
char ch;
int Find(int p){
    int cp=p;
    while(cp!=per[cp]){
        cp=per[cp];
    }
    int i=p,j;
    while(i!=cp){
        j=per[i];
        per[i]=cp;
        i=j;
    }
    return cp;
}
void Union(int p,int q)
{
    int cp=Find(p);
    int cq=Find(q);
    if(cp!=cq){
        per[cq]=cp;
    }
}
int main()
{
    scanf("%d %d",&n,&d);
    for(int i=1;i<=n;i++)
        scanf("%d %d",&a[i].x,&a[i].y);
    for(int i=1;i<=n;i++){
        per[i]=i;
    }
    memset(vis,0,sizeof(vis));
    while(scanf("%c",&ch)!=EOF){
        if(ch=='O'){
            //输入一个
            scanf("%d",&A);
            vis[A]=1;
            for(int i=1;i<=n;i++){
                if(vis[i]&&per[i]!=per[A]){
                    dis=(a[A].x-a[i].x)*(a[A].x-a[i].x)+(a[A].y-a[i].y)*(a[A].y-a[i].y);
                    if(dis<=d*d)
                    {
                        Union(A,i);
                    }
                }
            }
        }
        if(ch=='S'){
            scanf("%d%d",&A,&B);
            if(Find(A)==Find(B)) printf("SUCCESS\n");
            else  printf("FAIL\n");
        }
    }
    return 0;
}