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原理可以看hihocoder上面的讲解,很清楚，不多说了。
模板抄lrj训练指南上面的。
/** Treap 实现 名次树 功能： 1.找到排名为k的元素 2.值为x的元素的名次 初始化：Node* root = NULL; */ #include <cstdlib> #include <cstdio> #include <cstring> #include <vector> using namespace std; struct Node { Node * ch[2]; // 0左子树 1右子树 int r; // 随机优先级 int v; // 值 int s; // 以s为根的子树的大小 Node(int v):v(v) { ch[0] = ch[1] = NULL; r = rand(); s = 1; } bool operator<(const Node& rhs) const { // 按随机优先级排序 return r < rhs.r; } int cmp(int x) const { if (x == v) return -1; return x < v ? 0 : 1; } void maintain() // 更新 { s = 1; if (ch[0] != NULL) s += ch[0]->s; if (ch[1] != NULL) s += ch[1]->s; } } ; void rotate(Node* &o, int d) // d=0 代表左转, d=1代表右转 { Node* k = o->ch[d^1]; o->ch[d^1] = k->ch[d]; k->ch[d] = o; o->maintain(); k->maintain(); o = k; } void insert(Node* &o, int x) // o是根 x是插入的值 { if (o == NULL) { o = new Node(x); } else { int d = (x < o->v ? 0 : 1); // 不用cmp函数因为可能有重复的值 insert(o->ch[d], x); if ((o->ch[d]->r) > (o->r)) rotate(o, d^1); } o->maintain(); } void remove(Node* &o, int x) { int d = o->cmp(x); if (d == -1) { Node* u = o; if (o->ch[0] != NULL && o->ch[1] != NULL) { int d2 = ((o->ch[0]->r) > (o->ch[1]->r) ? 1 : 0); rotate(o, d2); remove(o->ch[d2], x); } else { if (o->ch[0] == NULL) o = o->ch[1]; else o = o->ch[0]; delete u; } } else { remove(o->ch[d], x); } if (o != NULL) o->maintain(); } int find(Node* o, int x) // 因为remove和insert都没有查值存不存在 记得操作之前调用find { while (o != NULL) { int d = o->cmp(x); if (d == -1) return 1; else o = o->ch[d]; } return 0; } int kth(Node* &o, int k, int fg) // fg=1第k大的值 fg=0第k小的值 返回0表示没找到 { if (o == NULL || k <= 0 || k > o->s) return 0; int s = (o->ch[fg] == NULL ? 0 : o->ch[fg]->s); if (k == s+1) return o->v; else if (k <= s) return kth(o->ch[fg], k, fg); else return kth(o->ch[fg^1], k-s-1, fg); } int prv(Node* &o, int x) // 查找x前面的元素 （<x的最大值 { if (o == NULL) return -1; if (x <= o->v) return prv(o->ch[0], x); int ans = prv(o->ch[1], x); return ans == -1 ? o->v : ans; } int nxt(Node* &o, int x) // 查找x后面的元素 （>x的最大值 { if (o == NULL) return -1; if (x >= o->v) return nxt(o->ch[1], x); int ans = nxt(o->ch[0], x); return ans == -1 ? o->v : ans; } void mergeto(Node* &src, Node* &dest) // 合并两棵树 把src加到dest上 src和dest都是树的根 { if (src->ch[0] != NULL) mergeto(src->ch[0], dest); if (src->ch[0] != NULL) mergeto(src->ch[1], dest); insert(dest, src->v); delete src; src = NULL; } void print(Node* &o) { if (o == NULL) return ; print(o->ch[0]); printf("%d ", o->v); print(o->ch[1]); } int main() { Node* root = NULL; insert(root, 3); insert(root, 3); insert(root, 4); insert(root, 5); remove(root, 4); print(root); return 0; }
例题：
上面hihocoder的例题，这个代码是照着讲解自己写的
//Treap.cpp #include <stdio.h> #include <string.h> #include <stdlib.h> const int N = 100005; struct Treap { int father, left, right; int key, weight; void init(int k, int w, int fa) { left = right = -1; father = fa, key = k, weight = w; } } tp[N]; int root; int treap_cnt; int new_treap(int k, int w, int fa = -1) { tp[treap_cnt].init(k, w, fa); return treap_cnt++; } void left_rotate(int a) // 左旋 把节点A的右儿子节点B转到A的父亲节点 { int b = tp[a].right; tp[b].father = tp[a].father; if (tp[tp[a].father].left == a) { // 判断a是父节点的左儿子还是右儿子 并用b替换 tp[tp[a].father].left = b; } else { tp[tp[a].father].right = b; } tp[a].right = tp[b].left; if (tp[b].left != -1) tp[tp[b].left].father = a; tp[b].left = a; tp[a].father = b; } void right_rotate(int a) // 右旋 把节点A的左儿子节点B转到A的父亲节点 { int b = tp[a].left; tp[b].father = tp[a].father; if (tp[tp[a].father].left == a) tp[tp[a].father].left = b; else tp[tp[a].father].right = b; tp[a].left = tp[b].right; if (tp[b].right != -1) tp[tp[b].right].father = a; tp[b].right = a; tp[a].father = b; } int insert(int a, int key) { if (key < tp[a].key) { if (tp[a].left == -1) { tp[a].left = new_treap(key, rand(), a); return tp[a].left; } else { return insert(tp[a].left, key); } } else { if (tp[a].right == -1) { tp[a].right = new_treap(key, rand(), a); return tp[a].right; } else { return insert(tp[a].right, key); } } } void rotate(int a) // 维持小顶堆 { int fa = tp[a].father; while (fa != -1) { if (tp[a].weight < tp[fa].weight) { if (a == tp[fa].left) right_rotate(fa); else left_rotate(fa); fa = tp[a].father; } else { break; } } if (fa == -1) root = a; } int find(int a, int key) { int cur = a, pre = -1; while (cur != -1) { if (tp[cur].key > key) { pre = cur; cur = tp[cur].left; } else if (tp[cur].key < key) { pre = cur; cur = tp[cur].right; } else { return key; } } while (pre != -1) { if (tp[pre].key < key) return tp[pre].key; pre = tp[pre].father; } return -2; } void print(int a) { if (a == -1) return; print(tp[a].left); printf("%d(%d) ", a, tp[a].key); print(tp[a].right); } int main(int argc, char const *argv[]) { //freopen("in", "r", stdin); root = -1; treap_cnt = 0; int n; char op[2]; int k; scanf("%d", &n); while (n--) { scanf("%s%d", op, &k); if (*op == ‘I‘) { if (root == -1) root = new_treap(k, rand()); else rotate(insert(root, k)); } else { printf("%d\n", find(root, k)); } } return 0; }
poj1442，直接套模板，比较简单
/** Treap 实现 名次树 功能： 1.找到排名为k的元素 2.值为x的元素的名次 */ #include <cstdlib> #include <cstdio> #include <cstring> #include <vector> using namespace std; struct Node { Node * ch[2]; // 0左子树 1右子树 int r; // 随机优先级 int v; // 值 int s; // 以s为根的子树的大小 Node(int v):v(v) { ch[0] = ch[1] = NULL; r = rand(); s = 1; } bool operator<(const Node& rhs) const { // 按随机优先级排序 return r < rhs.r; } int cmp(int x) const { if (x == v) return -1; return x < v ? 0 : 1; } void maintain() // 更新 { s = 1; if (ch[0] != NULL) s += ch[0]->s; if (ch[1] != NULL) s += ch[1]->s; } } ; void rotate(Node* &o, int d) // d=0 代表左转, d=1代表右转 { Node* k = o->ch[d^1]; o->ch[d^1] = k->ch[d]; k->ch[d] = o; o->maintain(); k->maintain(); o = k; } void insert(Node* &o, int x) // o是根 x是插入的值 { if (o == NULL) { o = new Node(x); } else { int d = (x < o->v ? 0 : 1); // 不用cmp函数因为可能有重复的值 insert(o->ch[d], x); if ((o->ch[d]->r) > (o->r)) rotate(o, d^1); } o->maintain(); } void remove(Node* &o, int x) { int d = o->cmp(x); if (d == -1) { Node* u = o; if (o->ch[0] != NULL && o->ch[1] != NULL) { int d2 = ((o->ch[0]->r) > (o->ch[1]->r) ? 1 : 0); rotate(o, d2); remove(o->ch[d2], x); } else { if (o->ch[0] == NULL) o = o->ch[1]; else o = o->ch[0]; delete u; } } else { remove(o->ch[d], x); } if (o != NULL) o->maintain(); } int find(Node* &o, int x) // 因为remove和insert都没有查值存不存在 记得操作之前调用find { while (o != NULL) { int d = o->cmp(x); if (d == -1) return 1; // exist else o = o->ch[d]; } return 0; // not exist } int kth(Node* &o, int k, int fg) // fg=1第k大的值 fg=0第k小的值 返回0表示没找到 { if (o == NULL || k <= 0 || k > o->s) return 0; int s = (o->ch[fg] == NULL ? 0 : o->ch[fg]->s); if (k == s+1) return o->v; else if (k <= s) return kth(o->ch[fg], k, fg); else return kth(o->ch[fg^1], k-s-1, fg); } void mergeto(Node* &src, Node* &dest) { if (src->ch[0] != NULL) mergeto(src->ch[0], dest); if (src->ch[0] != NULL) mergeto(src->ch[1], dest); insert(dest, src->v); delete src; src = NULL; } const int N = 30005; int a[N]; int main() { //freopen("in", "r", stdin); int m, n; while (~scanf("%d%d", &m, &n)) { Node* root = NULL; for (int i = 1; i <= m; ++i) { scanf("%d", a+i); } int x = 0, u, cnt = 0; for (int i = 0; i < n; ++i) { scanf("%d", &u); while (cnt < u) { insert(root, a[++cnt]); } printf("%d\n", kth(root, ++x, 0)); } } return 0; }
hdu4557
每个结点加了一个元素t，排序查找都要考虑t
#include <iostream> #include <cstdlib> #include <cstdio> #include <cstring> #include <map> #include <vector> using namespace std; struct Node { Node * ch[2]; // 0左子树 1右子树 int r; // 随机优先级 int v; // 值 int t; int s; // 以s为根的子树的大小 Node(int v, int t):v(v), t(t) { ch[0] = ch[1] = NULL; r = rand(); s = 1; } bool operator<(const Node& rhs) const { // 按随机优先级排序 if (r == rhs.r) return t < rhs.t; return r < rhs.r; } int cmp(int x, int y) const { if (x == v && y == t) return -1; if (x == v) return y < t ? 0 : 1; return x < v ? 0 : 1; } void maintain() // 更新 { s = 1; if (ch[0] != NULL) s += ch[0]->s; if (ch[1] != NULL) s += ch[1]->s; } } ; void rotate(Node* &o, int d) // d=0 代表左转, d=1代表右转 { Node* k = o->ch[d^1]; o->ch[d^1] = k->ch[d]; k->ch[d] = o; o->maintain(); k->maintain(); o = k; } void insert(Node* &o, int x, int y) // o是根 x是插入的值 { if (o == NULL) { o = new Node(x, y); } else { int d = o->cmp(x, y); insert(o->ch[d], x, y); if ((o->ch[d]->r) > (o->r)) rotate(o, d^1); } o->maintain(); } void remove(Node* &o, int x, int y) { int d = o->cmp(x, y); if (d == -1) { Node* u = o; if (o->ch[0] != NULL && o->ch[1] != NULL) { int d2 = ((o->ch[0]->r) > (o->ch[1]->r) ? 1 : 0); rotate(o, d2); remove(o->ch[d2], x, y); } else { if (o->ch[0] == NULL) o = o->ch[1]; else o = o->ch[0]; delete u; } } else { remove(o->ch[d], x, y); } if (o != NULL) o->maintain(); } int nxt(Node* &o, int x, int &res) // 查找x后面的元素 （>=x的最小值 { if (o == NULL) return -1; if (o->v < x) return nxt(o->ch[1], x, res); res = o->v; int ans = nxt(o->ch[0], x, res); return ans == -1 ? o->t : ans; } void print(Node* &o) { if (o == NULL) return ; print(o->ch[0]); printf("%d ", o->v); print(o->ch[1]); } int main() { freopen("in", "r", stdin); int T; cin >> T; int cas = 0; while (T--) { printf("Case #%d:\n", ++cas); int n; cin >> n; char op[10]; string name; int abi, res; Node* root = NULL; map<int, string> mp; for (int i = 0; i < n; ++i) { scanf("%s", op); if (*op == ‘A‘) { cin >> name; scanf("%d", &abi); mp[i] = name; insert(root, abi, i); printf("%d\n", root->s); } else { scanf("%d", &abi); int ans = nxt(root, abi, res); if (ans == -1) printf("WAIT...\n"); else { cout << mp[ans] << endl; remove(root, res, ans); } }//printf("debug:"); print(root); printf("\n"); } } }
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原文地址：http://www.cnblogs.com/wenruo/p/5762899.html
