poj2236,

## Problem:

Wireless Network
 Time Limit: 10000MS Memory Limit: 65536K Total Submissions: 9714 Accepted: 4110
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```
Source
POJ Monthly,HQM

## Solution:

Disjoint set structure, using find-union algorithms.
PS:
1st, take care of the union function, since the parent can be changed any time.
2nd, use the original index rather than using the root index. It won't make you efficient, it just make your code from AC to WA.

## Source Code:

//Sat Jun  5 11:26:21 CDT 2010
#include <vector>
#include <list>
#include <map>
#include <set>
#include <deque>
#include <queue>
#include <stack>
#include <bitset>
#include <algorithm>
#include <functional>
#include <numeric>
#include <utility>
#include <sstream>
#include <iostream>
#include <iomanip>
#include <cstdio>
#include <cmath>
#include <cstdlib>
#include <cctype>
#include <string>
#include <cstring>
#include <cstdio>
#include <cmath>
#include <cstdlib>
#include <ctime>
#define MMAX 1002
using namespace std;

class Networks
{
public:
int x;
int y;
int parent;
int rank;
bool visited;
};

vector<Networks> v(MMAX);

void Makeset(int N)
{
for (int i = 1; i <= N; i++)
{
int x, y;
cin >> x >> y;
v[i].x = x;
v[i].y = y;
v[i].parent = i;
v[i].rank = 1;
v[i].visited = false;
}
}

int Find(int x)
{
if (x != v[x].parent)
v[x].parent = Find(v[x].parent);
return v[x].parent;
}

void Union(int x, int y)
{
int t1 = Find(x);
int t2 = Find(y);
if (t1 == t2)
return;
if (v[t1].rank > v[t2].rank)
v[t2].parent = t1;
else
{
v[t1].parent = t2;
if (v[t1].rank == v[t2].rank)
v[t2].rank++;
}
}

bool check(int a, int b, long dist)
{
double d = (v[a].x - v[b].x) * (v[a].x - v[b].x) + (v[a].y - v[b].y)
* (v[a].y - v[b].y);
d = sqrt(d);
return d <= dist ? true : false;
}

int main(int argc, const char* argv[])
{
freopen("input.in", "r", stdin);
freopen("output.out", "w", stdout);
int N;
long D;
cin >> N >> D;
Makeset(N);
string c;
int a, b;
while (cin >> c >> a)
{
if (c == 'O')
{
for (int i = 1; i <= N; i++)
{
if (v[i].visited && i != a && check(i, a, D))
{
Union(i, a);
}
}
v[a].visited = true;
}
else
{
cin >> b;
int t1 = Find(a);
int t2 = Find(b);
if (check(a, b, D) || t1 == t2)
{

cout << "SUCCESS" << endl;
//              Union(a, b);
}
else
cout << "FAIL" << endl;
}
}
fclose(stdin);
fclose(stdout);
return 0;
}