Problem
有一个数列，从1排列到n，然后有Q个操作
- Top x:将第x个数放到序列的最前面
- Query x:询问x这个数在第几位
- Rank x:询问第x位数是什么
Solution
n非常的大，需要离散化：读入的Query操作和Top操作需要离散化
然后每当处理一个数时，用二分计算出离散化后的结果
对于Top操作，先把那个数删掉，然后加在splay的最左边。
Notice
离散化非常复杂
Code
#include<cmath>
#include<cstdio>
#include<cstring>
#include<iostream>
#include<algorithm>
using namespace std;
#define sqz main
#define ll long long
#define reg register int
#define rep(i, a, b) for (reg i = a; i <= b; i++)
#define per(i, a, b) for (reg i = a; i >= b; i--)
#define travel(i, u) for (reg i = head[u]; i; i = edge[i].next)
const int INF = 1e9, N = 500000;
const double eps = 1e-6, phi = acos(-1);
ll mod(ll a, ll b) {if (a >= b || a < 0) a %= b; if (a < 0) a += b; return a;}
ll read(){ ll x = 0; int zf = 1; char ch; while (ch != '-' && (ch < '0' || ch > '9')) ch = getchar();
if (ch == '-') zf = -1, ch = getchar(); while (ch >= '0' && ch <= '9') x = x * 10 + ch - '0', ch = getchar(); return x * zf;}
void write(ll y) { if (y < 0) putchar('-'), y = -y; if (y > 9) write(y / 10); putchar(y % 10 + '0');}
int s[N + 5], e[N + 5], point = 0, root, now, id[N + 5];
struct node
{
int val[N + 5], Size[N + 5], num[N + 5], son[2][N + 5], parent[N + 5];
inline void up(int u)
{
Size[u] = Size[son[0][u]] + Size[son[1][u]] + num[u];
}
void Newnode(int &u, int from, int v)
{
u = ++point;
parent[u] = from, son[0][u] = son[1][u] = 0;
num[u] = Size[u] = e[v] - s[v] + 1;
val[u] = v, id[v] = u;
}
void Build(int &u, int l, int r, int from)
{
int mid = (l + r) >> 1;
Newnode(u, from, mid);
if (l < mid) Build(son[0][u], l, mid - 1, u);
if (mid < r) Build(son[1][u], mid + 1, r, u);
up(u);
}
int Find(int x)
{
int l = 1, r = now;
while (l <= r)
{
int mid = (l + r) >> 1;
if (x >= s[mid] && x <= e[mid]) return mid;
else if (x < s[mid]) r = mid - 1;
else l = mid + 1;
}
}
void Rotate(int x, int &rt)
{
int y = parent[x], z = parent[y];
int l = (son[1][y] == x), r = 1 - l;
if (y == rt) rt = x;
else if (son[0][z] == y) son[0][z] = x;
else son[1][z] = x;
parent[x] = z;
parent[son[r][x]] = y, son[l][y] = son[r][x];
parent[y] = x, son[r][x] = y;
up(y);
up(x);
}
void Splay(int x, int &rt)
{
while (x != rt)
{
int y = parent[x], z = parent[y];
if (y != rt)
{
if ((son[0][z] == y) ^ (son[0][y] == x))
Rotate(x, rt);
else Rotate(y, rt);
}
Rotate(x, rt);
}
}
void Insert(int &u, int x, int last)
{
if (u == 0)
{
Newnode(u, last, x);
return;
}
else Insert(son[0][u], x, u);
up(u);
}
void Delete(int x)
{
Splay(x, root);
if (son[0][x] * son[1][x] == 0) root = son[0][x] + son[1][x];
else
{
int t = son[1][x];
while (son[0][t] != 0) t = son[0][t];
Splay(t, root);
son[0][t] = son[0][x], parent[son[0][x]] = t;
up(t);
}
parent[root] = 0;
}
int Find_rank(int x)
{
int t = id[Find(x)];
Splay(t, root);
return Size[son[0][root]] + 1;
}
int Find_num(int u, int k)
{
if (k <= Size[son[0][u]]) return Find_num(son[0][u], k);
else if (k <= Size[son[0][u]] + num[u]) return s[val[u]] + k - Size[son[0][u]] - 1;
else return Find_num(son[1][u], k - Size[son[0][u]] - num[u]);
}
void Top(int x)
{
int t = Find(x);
int y = id[t];
Delete(y);
Insert(root, t, 0);
Splay(point, root);
}
}Splay_tree;
int Q[N + 5], T[N + 5];
char st[N + 5][10];
int sqz()
{
int H_H = read();
rep(cas, 1, H_H)
{
int n = read(), q = read(), num = 0;
Q[0] = 0;
rep(i, 1, q)
{
scanf("%s%d", st[i], &T[i]);
if (st[i][0] == 'T' || st[i][0] == 'Q') Q[++num] = T[i];
}
Q[++num] = n;
sort(Q + 1, Q + num + 1);
now = 0;
rep(i, 1, num)
{
if (Q[i] == Q[i - 1]) continue;
if (Q[i] - Q[i - 1] > 1)
{
s[++now] = Q[i - 1] + 1;
e[now] = Q[i] - 1;
}
s[++now] = e[now] = Q[i];
}
point = 0;
Splay_tree.son[0][0] = Splay_tree.son[1][0] = Splay_tree.parent[0] = Splay_tree.Size[0] = Splay_tree.val[0] = Splay_tree.num[0] = 0;
Splay_tree.Build(root, 1, now, 0);
printf("Case %d:\n", cas);
rep(i, 1, q)
if (st[i][0] == 'T') Splay_tree.Top(T[i]);
else if (st[i][0] == 'Q') printf("%d\n", Splay_tree.Find_rank(T[i]));
else printf("%d\n", Splay_tree.Find_num(root, T[i]));
}
}
