POJ2236 Wireless Network（并查集）

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

好久没有写并查集了，做了个水题回忆一下

题目大意：n个点，给出每个点的坐标，以及距离上限d，有两种操作，‘O’操作激活一个点，‘s’操作查询两个点是否可达

可达：两个点距离小于d，或两个点可达同一个点

并查集将点并在一块，激活一个点就并一个点，并不了就先放在旁边，由于n最大1001，所以可以满足n方的复杂度进行点的搜索

AC代码

#include<iostream>
#include<cstdio>
#include<cmath>
using namespace std;
int fa[1010];
struct node
{
    double x,y,id;
}num[1010],temp[1010];
double dis(node a,node b)
{
    return sqrt((a.x-b.x)*(a.x-b.x)+(a.y-b.y)*(a.y-b.y));
}
void init(int x)
{
    for(int i=0;i<x+5;i++)
        fa[i]=i;
}
int ffind(int x)
{
    int t=x,r;
    while(t!=fa[t])
    {
        t=fa[t];
    }
    r=t;
    while(fa[x]!=r)
    {
        t=x;
        x=fa[x];
        fa[t]=r;
    }
    return r;
}
void Merge(int x,int y)
{
    int fx=ffind(x);
    int fy=ffind(y);
    if(fx!=fy)
    {
        fa[fx]=fy;
    }
}
int main()
{
    int n,d,p,q;
    int cnt=0;
    bool label;
    char opt[2];
    scanf("%d%d",&n,&d);
    init(n);
    for(int i=1;i<=n;i++)scanf("%lf%lf",&num[i].x,&num[i].y),num[i].id=i;
    while(~scanf("%s",opt))
    {
        if(opt[0]=='O')
        {
            label=false;
            scanf("%d",&p);
            for(int i=0;i<cnt;i++)
            {
                if(dis(temp[i],num[p])<=d)
                {
                    Merge(p,temp[i].id);
                    label=true;
                }
            }
            temp[cnt++]=num[p];
        }
        else
        {
            scanf("%d%d",&p,&q);
            int fx=ffind(p);
            int fy=ffind(q);
            if(fx==fy)
                printf("SUCCESS\n");
            else
                printf("FAIL\n");
        }
    }

    return 0;
}