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#include #include #include #include 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])); }}