poj 2236 Wireless Network

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
这题是并查集的应用，有一点点小小的变动。合并操作之前，要判断他们之间的距离是否小于d还要判断这两台电脑是否是好的。

#include<stdio.h>
#include<math.h>
int father[1002],rank[1002];//father[x]保存x的父节点，rank[x]记录x是否是好的
int a[1002][2];//保存各个点的坐标
int b[1002][1002];//记录各个点之间的距离
int find(int x)
{
    if(x!=father[x])
        father[x]=find(father[x]);
    return father[x];
}
void uion(int x,int y)
{
    x=find(x);
    y=find(y);
    if(x==y) return ;
    father[y]=x;    
}
int main()
{
    int n,d,i,j,f,e,sum;
    char c;
    scanf("%d%d",&n,&d);
    for(i=1;i<=n;i++)//初始化
    {
        father[i]=i;
        rank[i]=0;
    }
    sum=d*d;
    for(i=1;i<=n;i++)
        scanf("%d%d",&a[i][0],&a[i][1]);
    for(i=1;i<=n;i++)//算出各个点之间的距离
        for(j=1;j<=n;j++)
            b[i][j]=(a[i][0]-a[j][0])*(a[i][0]-a[j][0])+(a[i][1]-a[j][1])*(a[i][1]-a[j][1]);
    getchar();
    while(scanf("%c",&c)!=EOF)
    {
    
        if(c=='O')
        {
            scanf("%d",&e);
            rank[e]=1;
            for(i=1;i<=n;i++)
            {
                if(rank[i]==1&&b[e][i]<=sum&&i!=e)//满足条件的点进行合并
                    uion(e,i);
            }
        }
        else {
            scanf("%d%d",&e,&f);
            if(find(e)==find(f))//寻找父节点，并判断它们是否相等
                printf("SUCCESS
");
            else printf("FAIL
");
        }
        getchar();
    }
    return 0;
}