Wireless Network

题意：就是几台电脑，知道他们的坐标，两台修好的电脑如果距离<=d就可以联网， O p 代表修理p电脑 S p q代表链接pq

题解：并查集

坑点： FAIL 不要打成 FALL 我wa了的时候没找到原因，后来还是我看到别人的提醒才发现我也打成了FALL o(╯□╰)o

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

  1 #include<stdio.h>
  2 
  3 #include<iostream>
  4 
  5 #include<cmath>
  6 
  7 using namespace std;
  8 
  9 
 10 
 11 #define n 1010
 12 
 13 
 14 
 15 int pre[n],x[n],y[n];
 16 
 17 int m,d,k1,k2,i,r,r1;
 18 
 19 bool v[n];
 20 
 21 char ch;
 22 
 23 
 24 
 25 int find(int x)
 26 
 27 {
 28 
 29 if(x!=pre[x])
 30 
 31 pre[x]=find(pre[x]);
 32 
 33 return pre[x];
 34 
 35 }
 36 
 37 
 38 
 39 bool dis(int a,int b)//判断两台电脑之间的距离可不可以联网
 40 
 41 {
 42 
 43 int q=x[a]-x[b];
 44 
 45 int p=y[a]-y[b];
 46 
 47 if(q*q+p*p<=d*d)
 48 
 49 return true;
 50 
 51 else
 52 
 53 return false;
 54 
 55 }
 56 
 57 
 58 
 59 int main()
 60 
 61 {
 62 
 63 scanf("%d %d",&m,&d);
 64 
 65 for(i=1;i<=n;i++)//初始化
 66 
 67 {
 68 
 69 pre[i]=i;
 70 
 71 v[i]=false;
 72 
 73 }
 74 
 75 
 76 
 77 for(i=1;i<=m;i++)
 78 
 79 scanf("%d %d",&x[i],&y[i]);
 80 
 81 while(cin>>ch)
 82 
 83 {
 84 
 85 if(ch=='O')
 86 
 87 {
 88 
 89 scanf("%d",&r);
 90 
 91 v[r]=true;
 92 
 93 for(i=1;i<=m;i++)
 94 
 95 if(i!=r&&v[i]&&dis(i,r))
 96 
 97 {
 98 
 99 k1=find(r);
100 
101 k2=find(i);
102 
103 pre[k1]=k2;
104 
105 }
106 
107 }
108 
109 if(ch=='S')
110 
111 {
112 
113 scanf("%d %d",&r,&r1);
114 
115 k1=find(r);
116 
117 k2=find(r1);
118 
119 if(k1==k2)
120 
121 puts("SUCCESS");
122 
123 else
124 
125 puts("FAIL");
126 
127 }
128 
129 }
130 
131 return 0;
132 
133 }