平衡樹模板三題
阿新 • • 發佈：2019-02-07
您需要寫一種資料結構（可參考題目標題），來維護一個序列，其中需要提供以下操作：翻轉一個區間，例如原有序序列是 5 4 3 2 1，翻轉區間是 [2,4] 的話，結果是 5 2 3 4 1。
支援reverse（交換左右子樹）的Splay。
C++程式碼
#include <cstdio> template <class T> inline void swap(T &a, T &b) { register T c; c = a; a = b; b = c; } const int mxn = 1000005; int n, m; int root; int fat[mxn]; int lsn[mxn]; int rsn[mxn]; int rev[mxn]; int siz[mxn]; int build(int l, int r) { int t = (l + r) >> 1; if (l < t)fat[lsn[t] = build(l, t - 1)] = t; if (r > t)fat[rsn[t] = build(t + 1, r)] = t; siz[t] = siz[lsn[t]] + siz[rsn[t]] + 1; return t; } inline void update(int t) { siz[t] = siz[lsn[t]] + siz[rsn[t]] + 1; } inline void pushdown(int t) { if (t && rev[t]) { rev[t] = 0; swap(lsn[t], rsn[t]); if (lsn[t])rev[lsn[t]] ^= 1; if (rsn[t])rev[rsn[t]] ^= 1; } } inline void pushdown(int t, int p) { static int stk[mxn], top; for (stk[++top] = t; t != p; ) stk[++top] = t = fat[t]; while (top)pushdown(stk[top--]); } inline void rotate(int t) { int f = fat[t]; int g = fat[f]; if (t == lsn[f]) { lsn[f] = rsn[t]; if (rsn[t]) fat[rsn[t]] = f; rsn[t] = f; fat[f] = t; fat[t] = g; if (g) { if (lsn[g] == f) lsn[g] = t; else rsn[g] = t; } else root = t; } else { rsn[f] = lsn[t]; if (lsn[t]) fat[lsn[t]] = f; lsn[t] = f; fat[f] = t; fat[t] = g; if (g) { if (lsn[g] == f) lsn[g] = t; else rsn[g] = t; } else root = t; } update(f); update(t); } inline void splay(int t, int p) { pushdown(t, p); while (fat[t] != p) { int f = fat[t]; int g = fat[f]; if (g == p) rotate(t); else { int a = lsn[f] == t; int b = lsn[g] == f; if (a == b) rotate(f); else rotate(t); rotate(t); } } } inline int kth(int k) { for (int t = root; t; ) { pushdown(t); int s = siz[lsn[t]]; if (k <= s) t = lsn[t]; else if (k == ++s) return t; else t = rsn[t], k -= s; } return 0; } void travel(int t) { pushdown(t); if (lsn[t])travel(lsn[t]); if (t > 1 && t < n + 2) printf("%d ", t - 1); if (rsn[t])travel(rsn[t]); } signed main() { scanf("%d%d", &n, &m); root = build(1, n + 2); for (int i = 1, l, r; i <= m; ++i) { scanf("%d%d", &l, &r), ++l, ++r; int a = kth(l - 1); int b = kth(r + 1); splay(a, 0); splay(b, a); rev[lsn[b]] ^= 1; } travel(root); }
二逼平衡樹
您需要寫一種資料結構（可參考題目標題），來維護一個有序數列，其中需要提供以下操作：
- 查詢 x 在區間內的排名；
- 查詢區間內排名為 k 的值；
- 修改某一位置上的數值；
- 查詢 x 在區間內的前趨（前趨定義為小於 x，且最大的數）；
- 查詢 x 在區間內的後繼（後繼定義為大於 x，且最小的數）。
C++程式碼
#include <cstdio> #define min(a, b) (a < b ? a : b) #define max(a, b) (a > b ? a : b) template <class T> inline T getmin(const T &a, const T &b) { return min(a, b); } template <class T> inline T getmax(const T &a, const T &b) { return max(a, b); } const int mxn = 100005; const int mxm = 5000005; const int inf = 100000008; namespace splay { int tot; int siz[mxm]; int num[mxm]; int fat[mxm]; int son[mxm][2]; int insert(int &t, int f, int v) { if (!t) { t = ++tot; fat[t] = f; num[t] = v; siz[t] = 1; son[t][0] = 0; son[t][1] = 0; return t; } ++siz[t]; if (v <= num[t]) return insert(son[t][0], t, v); else return insert(son[t][1], t, v); } #define update(t) (siz[t] = siz[son[t][0]] + siz[son[t][1]] + 1) inline void rotate(int t) { int f = fat[t]; int g = fat[f]; int a = son[f][1] == t; int b = !a, s = son[t][b]; fat[t] = g; fat[f] = t; son[t][b] = f; son[f][a] = s; if (s) fat[s] = f; if (g) { if (son[g][0] == f) son[g][0] = t; else son[g][1] = t; } update(f); update(t); } inline void splay(int t, int p = 0) { while (fat[t] != p) { int f = fat[t]; int g = fat[f]; if (g == p) rotate(t); else { int a = son[f][1] == t; int b = son[g][1] == f; if (a == b) rotate(f), rotate(t); else rotate(t), rotate(t); } } } inline int rnk(int t, int v) { int k = 0; while (t) { if (v <= num[t]) t = son[t][0]; else { k += siz[t]; t = son[t][1]; k -= siz[t]; } } return k; } inline int kth(int t, int k) { while (t) { int s = siz[son[t][0]]; if (k <= s) t = son[t][0]; else if (k == ++s) return t; else t = son[t][1], k -= s; } return 0; } inline int fnd(int t, int v) { while (t) { if (num[t] == v) return t; if (v < num[t]) t = son[t][0]; else t = son[t][1]; } return 0; } inline int pre(int t, int v) { int r = -inf; while (t) { if (v <= num[t]) t = son[t][0]; else { r = max(r, num[t]); t = son[t][1]; } } return r; } inline int nxt(int t, int v) { int r = +inf; while (t) { if (v >= num[t]) t = son[t][1]; else { r = min(r, num[t]); t = son[t][0]; } } return r; } inline void ins(int &t, int v) { int p = insert(t, 0, v); splay(p, 0); t = p; } inline void rmv(int &t, int v) { int p = fnd(t, v); if (p) { splay(p, 0); if (!son[p][0]) t = son[p][1], fat[t] = 0; else if (!son[p][1]) t = son[p][0], fat[t] = 0; else { t = son[p][0]; while (son[t][1]) t = son[t][1]; splay(t, p); son[t][1] = son[p][1]; fat[son[p][1]] = t; fat[t] = 0; update(t); } } } void print(int t) { if (!t)return; print(son[t][0]); printf("%d: [%d] (%d) (%d %d) {%d}\n", t, num[t], fat[t], son[t][0], son[t][1], siz[t]); print(son[t][1]); } #undef update } namespace stree { int tot; int rot[mxm]; int lsn[mxm]; int rsn[mxm]; int build(int l, int r, int *s) { int t = ++tot; for (int i = l; i <= r; ++i) splay::ins(rot[t], s[i]); if (l != r) { int m = (l + r) >> 1; lsn[t] = build(l, m, s); rsn[t] = build(m + 1, r, s); } return t; } void update(int t, int l, int r, int p, int a, int b) { splay::rmv(rot[t], a); splay::ins(rot[t], b); if (l != r) { int m = (l + r) >> 1; if (p <= m) update(lsn[t], l, m, p, a, b); else update(rsn[t], m + 1, r, p, a, b); } } int queryRnk(int t, int l, int r, int x, int y, int v) { if (l == x && y == r) return splay::rnk(rot[t], v); int m = (l + r) >> 1; if (y <= m) return queryRnk(lsn[t], l, m, x, y, v); else if (x > m) return queryRnk(rsn[t], m + 1, r, x, y, v); return queryRnk(lsn[t], l, m, x, m, v) + queryRnk(rsn[t], m + 1, r, m + 1, y, v); } int queryPre(int t, int l, int r, int x, int y, int v) { if (l == x && y == r) return splay::pre(rot[t], v); int m = (l + r) >> 1; if (y <= m) return queryPre(lsn[t], l, m, x, y, v); else if (x > m) return queryPre(rsn[t], m + 1, r, x, y, v); return getmax( queryPre(lsn[t], l, m, x, m, v), queryPre(rsn[t], m + 1, r, m + 1, y, v)); } int queryNxt(int t, int l, int r, int x, int y, int v) { if (l == x && y == r) return splay::nxt(rot[t], v); int m = (l + r) >> 1; if (y <= m) return queryNxt(lsn[t], l, m, x, y, v); else if (x > m) return queryNxt(rsn[t], m + 1, r, x, y, v); return getmin( queryNxt(lsn[t], l, m, x, m, v), queryNxt(rsn[t], m + 1, r, m + 1, y, v)); } } int n, m, s[mxn]; signed main() { // freopen("in", "r", stdin); // freopen("out", "w", stdout); scanf("%d%d", &n, &m); for (int i = 1; i <= n; ++i) scanf("%d", s + i); stree::build(1, n, s); for (int i = 1, k, x, y, z; i <= m; ++i) { scanf("%d%d%d", &k, &x, &y); if (k == 3) stree::update(1, 1, n, x, s[x], y), s[x] = y; else { scanf("%d", &z); if (k == 1) printf("%d\n", stree::queryRnk(1, 1, n, x, y, z) + 1); else if (k == 4) printf("%d\n", stree::queryPre(1, 1, n, x, y, z)); else if (k == 5) printf("%d\n", stree::queryNxt(1, 1, n, x, y, z)); else { int lt = -inf, rt = +inf, mid, ans; while (lt <= rt) { mid = (lt + rt) >> 1; if (stree::queryRnk(1, 1, n, x, y, mid) < z) lt = mid + 1, ans = mid; else rt = mid - 1; } printf("%d\n", ans); } } } }