poj2236(Wireless Network)并查集

4 1
0 1
0 2
0 3
0 4
O 1
O 2
O 4
S 1 4
O 3
S 1 4

题意：给你n个坏的电脑的坐标，如果两个修好的电脑距离不大于d，他们之间连通。两台电脑可以通过第三台电脑间接连通。给出若干次询问，一种是修哪台电脑，一种是询问两台电脑是否连通。

题解：典型的并查集，合并时我也没有加rank数组优化，可以自己加一下，不过10s时间绰绰有余了。

#include <iostream>
#include <sstream>
#include <fstream>
#include <string>
#include <map>
#include <vector>
#include <list>
#include <set>
#include <stack>
#include <queue>
#include <deque>
#include <algorithm>
#include <functional>
#include <iomanip>
#include <limits>
#include <new>
#include <utility>
#include <iterator>
#include <cstdio>
#include <cstdlib>
#include <cstring>
#include <cctype>
#include <cmath>
#include <ctime>
using namespace std;

const int maxn = 1010;

int f[maxn];
int px[maxn], py[maxn];
int vis[maxn];
int n, d;

int Find(int x)
{
    return x == f[x] ? x : (f[x] = Find(f[x]));
}

void join(int x, int y)
{
    int fx = Find(x), fy = Find(y);
    f[fx] = fy;
}

int dis(int x, int y)
{
    return (px[x]-px[y])*(px[x]-px[y])+(py[x]-py[y])*(py[x]-py[y]);
}

int main()
{
    cin >> n >> d;
    for (int i = 1; i <= n; ++i)
    {
        scanf("%d%d", &px[i], &py[i]);
        f[i] = i;
    }
    memset(vis, 0, sizeof(vis));
    char op[5];
    while (~scanf("%s", op))
        if (op[0] == 'O')
        {
            int p;
            scanf("%d", &p);
            vis[p] = 1;
            for (int i = 1; i <= n; ++i)
                if (i != p && vis[i] && dis(i, p) <= d*d)
                    join(i, p);
        }
        else
        {
            int x, y;
            scanf("%d%d", &x, &y);
            int fx = Find(x), fy = Find(y);
            if (fx == fy)
                printf("SUCCESS\n");
            else
                printf("FAIL\n");
        }
    return 0;
}