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 }