POJ2236

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

Sample Output

FAIL
SUCCESS

版本一：

使用数组

 1 #include<iostream>
 2 #include<stdio.h>
 3 #include<string.h>
 4 #include<algorithm>
 5 #include<cmath>
 6 #define maxn 50010
 7 using namespace std;
 8 int n;
 9 double d;
10 double dx[1005],dy[1005];
11 int pre[1005],repair[1005];
12 double dis(int a,int b)
13 {
14     return sqrt((dx[a]-dx[b])*(dx[a]-dx[b])+(dy[a]-dy[b])*(dy[a]-dy[b]));
15 }
16 int Find(int x)
17 {
18     if(x!=pre[x])
19         return pre[x]=Find(pre[x]);
20 }
21 int Union(int x,int y)
22 {
23     pre[Find(x)]=Find(y);
24 }
25 int main()
26 {
27     cin>>n>>d;
28     for(int i=1;i<=n;i++)
29     {
30         cin>>dx[i]>>dy[i];
31         pre[i]=i;
32     }
33     char k;
34     int p,q,len=0;
35     getchar();
36     while(~scanf("%c",&k))
37     {
38         if(k=='O')
39             {
40                 cin>>p;
41                 repair[len++]=p;
42                 for(int i=0;i<len-1;i++)
43                     if(dis(p,repair[i])<=d)
44                        Union(repair[i],p);
45             }
46             else
47             {
48                 cin>>p>>q;
49                 if(Find(p)==Find(q))
50                     cout<<"SUCCESS"<<endl;
51                 else
52                     cout<<"FAIL"<<endl;
53             }
54             getchar();
55     }
56     return 0;
57 }

版本二：

使用结构体

 1 #include<iostream>
 2 #include<stdio.h>
 3 #include<string.h>
 4 #include<algorithm>
 5 #include<cmath>
 6 #define maxn 1005
 7 using namespace std;
 8 int n;
 9 double d;
10 struct node
11 {
12     double x,y;
13     int pre;//根节点
14     int ranks;//树的高度
15     bool repair;//是否修复
16 }link[maxn];
17 int Find(int x)
18 {
19     if(x!=link[x].pre)
20     return link[x].pre=Find(link[x].pre);
21 }
22 double dis(int i,int j)
23 {
24     return sqrt((link[i].x-link[j].x)*(link[i].x-link[j].x)+(link[i].y-link[j].y)*(link[i].y-link[j].y));
25 }
26 void Union(int i,int j)
27 {
28     //if(link[i].repair&&link[j].repair)
29        // return 0;
30     int root1=Find(i),root2=Find(j);
31     //if(root1==root2)return 0;
32     if(link[root1].ranks>link[root2].ranks)
33     {
34         link[root2].pre=root1;
35         link[root1].ranks+=link[root2].ranks;
36     }
37     else
38     {
39         link[root1].pre=root2;
40         link[root2].ranks+=link[root1].ranks;
41     }
42     //return 1;
43 }
44 
45 
46 int main()
47 {
48     cin>>n>>d;
49     for(int i=1;i<=n;i++)
50     {
51         cin>>link[i].x>>link[i].y;
52         link[i].ranks=1;
53         link[i].pre=i;
54         link[i].repair=0;
55     }
56     char k;
57     int p,q;
58     getchar();
59     while(~scanf("%c",&k))
60     {
61         if(k=='O')
62             {
63                 cin>>p;
64                 link[p].repair=1;
65                 for(int i=1;i<=n;i++)
66                 {
67                     if(Find(i)!=Find(p)&&link[i].repair&&dis(i,p)<=d)
68                         Union(p,i);
69                 }
70             }
71             else
72             {
73                 cin>>p>>q;
74                 if(Find(p)==Find(q))
75                     cout<<"SUCCESS"<<endl;
76                 else
77                     cout<<"FAIL"<<endl;
78             }
79             getchar();
80     }
81     return 0;
82 }

题目大意：

n个站点，已知站点的坐标。

‘O'操作每次修通一个站点，该站点与距离该站点小于d且也为修通的站点联通；

’S'操作查询两个站点是否联通。

思路：

并查集的题。设置距离函数，如果满足距离条件则将两个点合并。如果查询到的两点根节点相同则能联通。