POJ 2236 Wireless Network

/* 此题是并查集的应用，只不过是增加了一个判断距离的
* 条件，这样每次输入一个点时，就需要和所有以前的点
* 进行判断，因此时间有点多，没有想到很好的方法，真不
* 知道100ms是怎么做出来的。。。。
* */

#include<iostream>
using namespace std;

struct Node{
    int x,y;
}node[1010];
int op[1010],tree[1010],n,d;
void make_set()
{
    for(int i=0;i<=n;++i)
        tree[i] = -1;
}
int f(const int& x,const int& y,const int& xx,const int& yy)
{
    int tmp = (x-xx)*(x-xx)+(y-yy)*(y-yy);
    return tmp>d?0:1;
}
int find_set(int x)
{
    int root = x;
    int tmp;
    while(tree[root]>=0)
        root = tree[root];
    while(x!=root)
    {
        tmp = tree[x];
        tree[x] = root;
        x = tmp;
    }
    return root;
}
void union_set(const int& root1,const int& root2)
{
    int a = tree[root1];
    int b = tree[root2];
    if(a>b)
    {
        tree[root1] = root2;
        tree[root2] = a+b;
    }else
    {
        tree[root2] = root1;
        tree[root1] = a+b;
    }
}
int main()
{
    int i,j,root1,root2,n1,n2,cnt=0;
    char ch;
    cin>>n>>d;
    d = d*d;
    for(i=1;i<=n;++i)
        scanf("%d%d",&node[i].x,&node[i].y);
        make_set();
    while(scanf("%c",&ch) != EOF)
    {
    
        if(ch == 'O')
        {
            scanf("%d",&op[cnt]);
            for(i=0;i<cnt;++i)
                if(f(node[op[cnt]].x,node[op[cnt]].y,node[op[i]].x,node[op[i]].y))
                {
                    root1 = find_set(op[cnt]);
                    root2 = find_set(op[i]);
                    if(root1!=root2)
                        union_set(root1,root2);
                }
                ++cnt;
        }else if(ch == 'S')
        {
            scanf("%d%d",&n1,&n2);
            root1 = find_set(n1);
            root2 = find_set(n2);
            if(root1!=root2)
                printf("FAIL\n");
            else
                printf("SUCCESS\n");
        }
    }
    return 0;
}