就是给一个数列，维护操作：（1）加一个数（2）求当前全部数的第K大。。。
看了Claris大爷的做法深有启发，于是本蒟蒻的替罪羊树的第一次就没了。。。
写完才发现。。。Orz主席树怎么忘了、、、貌似实现更简单啊！！！
不管了QAQ
1 /************************************************************** 2 Problem: 3570 3 User: rausen 4 Language: C++ 5 Result: Accepted 6 Time:2284 ms 7 Memory:6288 kb 8 ****************************************************************/ 9 10 #include <cstdio> 11 #include <cmath> 12 13 using namespace std; 14 typedef long long ll; 15 typedef double lf; 16 const lf A = 0.8; 17 const int N = 200005; 18 19 int n, C; 20 int size[N], son[N][2], val[N], f[N], tot, root; 21 int data[N], id[N], cnt; 22 23 inline int read() { 24 int x = 0; 25 char ch = getchar(); 26 while (ch < '0' || '9' < ch) 27 ch = getchar(); 28 while ('0' <= ch && ch <= '9') { 29 x = x * 10 + ch - '0'; 30 ch = getchar(); 31 } 32 return x; 33 } 34 35 int ins(int x, int p) { 36 ++size[x]; 37 int S = p >= val[x]; 38 if (!son[x][S]) { 39 son[x][S] = ++tot; 40 f[tot] = x, size[tot] = 1, val[tot] = p; 41 return tot; 42 }else return ins(son[x][S], p); 43 } 44 45 int build(int fa, int l, int r) { 46 int mid = l + r >> 1, x = id[mid]; 47 f[x] = fa, son[x][0] = son[x][1] = 0, size[x] = 1; 48 val[x] = data[mid]; 49 if (l == r) return x; 50 if (l < mid) size[x] += size[son[x][0] = build(x, l, mid - 1)]; 51 if (mid < r) size[x] += size[son[x][1] = build(x, mid + 1, r)]; 52 return x; 53 } 54 55 void dfs(int x) { 56 if (son[x][0]) dfs(son[x][0]); 57 data[++cnt] = val[x], id[cnt] = x; 58 if (son[x][1]) dfs(son[x][1]); 59 } 60 61 inline int rebuild(int x) { 62 cnt = 0; 63 dfs(x); 64 return build(f[x], 1, cnt); 65 } 66 67 inline void insert(int p) { 68 if (!root) { 69 root = tot = size[1] = 1; 70 val[1] = p; 71 return; 72 } 73 int x = ins(root, p), dep = 0, z = x; 74 while (f[z]) z = f[z], ++dep; 75 if (dep < log(tot) / log(1 / A)) return; 76 while ((lf) size[son[x][0]] < A * size[x] && (lf)size[son[x][1]] < A * size[x]) 77 x = f[x]; 78 if (!x) return; 79 if (x == root) { 80 root = rebuild(x); 81 return; 82 } 83 int y = f[x], S = son[y][1] == x; 84 son[y][S] = rebuild(x); 85 } 86 87 inline int find_kth(int k) { 88 int x = root, sum; 89 while (1) { 90 sum = size[son[x][0]] + 1; 91 if (k == sum) return val[x]; 92 if (k < sum) x = son[x][0]; 93 else k -= sum, x = son[x][1]; 94 } 95 } 96 97 int main() { 98 int x, y, z; 99 n = read(), C = read(); 100 while (n--) { 101 x = read(), read(), read(); 102 insert(x); 103 } 104 n = read(); 105 while (n--) { 106 x = read(); 107 if (x) { 108 y = read(), z = find_kth(read()); 109 printf("%.3lf ", sqrt((ll) 2 * C * y + (ll) z * z)); 110 } else 111 insert(read()), read(), read(); 112 } 113 return 0; 114 }