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：
1 #include<cstdio> 2 #include<cstring> 3 #include<algorithm> 4 #include<cmath> 5 using namespace std; 6 const int maxn = 1005; 7 int father[maxn], vis[maxn]; 8 9 struct node{ 10 double x, y; //此处用double 11 }p[maxn]; 12 13 int Find(int x) 14 { 15 return father[x] == x ? x : father[x] = Find(father[x]); 16 } 17 18 void Union(int x, int y) 19 { 20 int rootx = Find(x), rooty = Find(y); 21 if(rootx != rooty) father[rootx] = rooty; 22 return ; 23 } 24 25 int main() 26 { 27 int n, d, k, a, b; 28 char ch; 29 scanf("%d %d", &n, &d); 30 memset(vis, 0, sizeof(vis)); 31 for(int i = 1; i <= n; i++) father[i] = i; 32 for(int i = 1; i <= n; i++) scanf("%lf %lf", &p[i].x, &p[i].y); 33 while(~scanf(" %c", &ch)) 34 { 35 if(ch == 'O') 36 { 37 scanf("%d", &k); 38 vis[k] = 1; 39 for(int i = 1; i <= n; i++) 40 { 41 if(!vis[i]) continue; //必须是已经修过的 42 //两点距离小于等于d,则可以合并 43 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) 44 Union(i, k); 45 } 46 } 47 else 48 { 49 scanf("%d %d", &a, &b); 50 //根节点相同，则表明可以连接 51 if(Find(a) == Find(b)) printf("SUCCESS "); 52 else printf("FAIL "); 53 } 54 } 55 return 0; 56 }