A - Simple String Problem
Time Limit:10000MS Memory Limit:65536KB 64bit IO Format:%lld & %llu
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 //题意是，电脑都坏了，可以一个一个修复，电脑只能在一定的距离内才能连通，在询问是否连通的时候输出是否连通。第一行是 n ，d ，d 是代表电脑能连通的最大距离，然后是 n 行坐标，在然后是命令 O 代表修复电脑，S 代表查询两个电脑是否连通 并查集简单的应用，压缩了路径就只用了1秒1125 ms
1 #include <iostream>
2 #include <stdio.h>
3 #include <math.h>
4 using namespace std;
5
6 struct Com
7 {
8 int x,y;
9 int on;
10 }com[1005];
11 int p[1005];
12
13 int find(int x)
14 {
15 if (x!=p[x])
16 p[x]=find(p[x]);
17 return p[x];
18 }
19
20 int Distance(int a,int b,double x)
21 {
22 double j=(com[a].x-com[b].x)*(com[a].x-com[b].x);
23 double k=(com[a].y-com[b].y)*(com[a].y-com[b].y);
24 double l=sqrt(j+k);
25 if (x<l)
26 return 1;
27 return 0;
28 }
29
30 int main()
31 {
32 int n,i;
33 double s;
34 scanf("%d%lf",&n,&s);
35 for (i=1;i<=n;i++)
36 {
37 scanf("%d%d",&com[i].x,&com[i].y);
38 com[i].on=0;
39 p[i]=i;
40 }
41 char str[2];
42 int k;
43 while(scanf("%s",str)!=EOF)
44 {
45 if (str[0]==‘O‘)
46 {
47 scanf("%d",&k);
48 if (com[k].on==1)
49 continue;
50 com[k].on=1;
51 for (i=1;i<=n;i++)
52 {
53 if (i==k||com[i].on==0) //未修复
54 continue;
55 if (Distance(i,k,s))//距离超出
56 continue;
57 int fa=find(k);
58 int fb=find(i);
59 p[fa]=fb;
60 }
61 }
62 if (str[0]==‘S‘)
63 {
64 int a,b;
65 scanf("%d%d",&a,&b);
66 int fa=find(a),fb=find(b);
67 if (fa==fb)
68 {
69 printf("SUCCESS\n");
70 //没有这种情况
71 /* if (a!=b)
72 printf("SUCCESS\n");
73 else if (a==b&&com[a].on)
74 printf("SUCCESS\n");
75 else
76 printf("FAIL\n");*/
77 }
78 else
79 printf("FAIL\n");
80 }
81 }
82 return 0;
83 }