Wireless Network POJ

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：有n台损坏的计算机，计算机在d米内才能连接在一起，两个a和b能连接，b和c能连接，那么a和c能通过b而连接。
然后有两个操作，大写字母O和大写字母S，O表示将损坏的计算机修好，S表示测试a和b之间能否连接。输入第一行n,d.
然后n行表示第n个电脑的位置，再往下是语句，执行每个语句，对于测试语句要输出测试结果是SUCCESS还是FAIL。

思路：简单的并查集，先将计算机的坐标之间的关系转换成并查集中点与点之间的关系，最后判断即可。

代码：

#include <cstdio>
#include <fstream>
#include <algorithm>
#include <cmath>
#include <deque>
#include <vector>
#include <queue>
#include <string>
#include <cstring>
#include <map>
#include <stack>
#include <set>
#include <sstream>
#include <iostream>
#define mod 998244353
#define eps 1e-6
#define ll long long
#define INF 0x3f3f3f3f
using namespace std;

//记录计算机的坐标
struct node
{
    int x,y;
};
node no[1005];
//保存能与与第x个计算机相连的计算机
vector<int> ve[1005];
//保存该计算机是否被修复
bool vis[1005];

//fa[x]表示x的最远祖先
int fa[50002];
//初始化，一开始每个点单独成集合
void build(int qwq) 
{
    for(int i=1;i<=qwq;i++)
    {
        fa[i]=i;
    }
    return ;
} 
//找到x的最远祖先，并且压缩路径
int find(int x)
{
    if(fa[x]==x)
    {
        return x;
    }
    return fa[x]=find(fa[x]);
}
//判断x,y是不是在同一个集合里，直接判断最远祖先是不是一样的 
bool che(int x,int y)
{
    return find(x)==find(y);
}
//合并x,y，我们在判断x和y是不是同一个集合里，
//路径压缩之后fa[x],fa[y]已经是最远祖先了，
//所以直接将fa[x]的父亲连接在fa[y]的祖先上
void mer(int x,int y)
{
    if(!che(x,y)) 
    {
        fa[fa[x]]=fa[y];
    }
    return ;
}
int distan(node a,node b)
{
    return (a.x-b.x)*(a.x-b.x)+(a.y-b.y)*(a.y-b.y);
}
//修复计算机x并将集合x与和x相连的集合合并
void updata(int x)
{
    vis[x]=true;
    for(int i=0;i<ve[x].size();i++)
    {
        //与x相连的点
        int y=ve[x][i];
        //如果y点被唤醒则合并x与y
        if(vis[y]&&!che(x,y))
        {
            mer(x,y);
        }
    }
}
int main()
{
    int n,d;
    scanf("%d %d",&n,&d);
    //初始化信息
    build(n);
    //记录点的信息
    for(int i=1;i<=n;i++)
    {
        scanf("%d %d",&no[i].x,&no[i].y);
    }
    //遍历判断哪些点之间能连接在一起
    for(int i=1;i<=n;i++)
    {
        for(int j=i+1;j<=n;j++)
        {
            if(distan(no[i],no[j])<=d*d)
            {
                ve[i].push_back(j);
                ve[j].push_back(i);
            }
        }
    }
    char ch;
    int a,b;
    while(scanf("%c",&ch)!=EOF)
    {
        if(ch=='O')
        {
            scanf("%d",&a);
            updata(a);
        }
        else if(ch=='S')
        {
            scanf("%d %d",&a,&b);
            //当两个点的祖先一样时表示两点能连接在一起
            if(find(a)==find(b))
            {
                printf("SUCCESS
");
            }
            else
            {
                printf("FAIL
");
            }
        }
    }
}