POJ2236 Wireless Network

解题思路：简单并查集，注意时间限制是10000MS，每次进行O操作之后，

　　　　  进行一次for循环，进行相关调整。同时注意输入输出格式，见代码：

 #include<cstdio>
 #include<cstring>
 #include<algorithm>
 #include<cmath>
 using namespace std;
 const int maxn = ;
 int father[maxn], vis[maxn];

 struct node{
     double x, y; //此处用double
 }p[maxn];

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

 void Union(int x, int y)
 {
     int rootx = Find(x), rooty = Find(y);
     if(rootx != rooty) father[rootx] = rooty;
     return ;
 }

 int main()
 {
     int n, d, k, a, b;
     char ch;
     scanf("%d %d", &n, &d);
     memset(vis, , sizeof(vis));
     for(int i = ; i <= n; i++) father[i] = i;
     for(int i = ; i <= n; i++) scanf("%lf %lf", &p[i].x, &p[i].y);
     while(~scanf(" %c", &ch))
     {
         if(ch == 'O')
         {
             scanf("%d", &k);
             vis[k] = ;
             for(int i = ; i <= n; i++)
             {
                 if(!vis[i]) continue; //必须是已经修过的
                 //两点距离小于等于d,则可以合并
                 if(sqrt((p[i].x-p[k].x)*(p[i].x-p[k].x)+(p[i].y-p[k].y)*(p[i].y-p[k].y)) <= d)
                 Union(i, k);
             }
         }
         else
         {
             scanf("%d %d", &a, &b);
             //根节点相同，则表明可以连接
             if(Find(a) == Find(b)) printf("SUCCESS\n");
             else printf("FAIL\n");
         }
     }
     return ;
 }