[BZOJ3224]Tyvj 1728 普通平衡树
试题描述
您需要写一种数据结构（可参考题目标题），来维护一些数，其中需要提供以下操作：
1. 插入x数
2. 删除x数(若有多个相同的数，因只删除一个)
3. 查询x数的排名(若有多个相同的数，因输出最小的排名)
4. 查询排名为x的数
5. 求x的前驱(前驱定义为小于x，且最大的数)
6. 求x的后继(后继定义为大于x，且最小的数)
输入
第一行为n，表示操作的个数,下面n行每行有两个数opt和x，opt表示操作的序号(1<=opt<=6)
输出
对于操作3,4,5,6每行输出一个数，表示对应答案
输入示例
10 1 106465 4 1 1 317721 1 460929 1 644985 1 84185 1 89851 6 81968 1 492737 5 493598
输出示例
106465 84185 492737
数据规模及约定
1.n的数据范围：n<=100000
2.每个数的数据范围：[-1e7,1e7]
题解
treap 模板题。
#include <iostream>
#include <cstdio>
#include <cstring>
#include <cstdlib>
#include <cctype>
#include <algorithm>
using namespace std;
int read() {
int x = 0, f = 1; char c = getchar();
while(!isdigit(c)){ if(c == '-') f = -1; c = getchar(); }
while(isdigit(c)){ x = x * 10 + c - '0'; c = getchar(); }
return x * f;
}
#define maxn 100010
struct Node {
int v, r, siz;
Node() {}
Node(int _, int __): v(_), r(__) {}
} ns[maxn];
int rt, ToT, fa[maxn], ch[2][maxn];
void maintain(int o) {
ns[o].siz = 1;
for(int i = 0; i < 2; i++) if(ch[i][o])
ns[o].siz += ns[ch[i][o]].siz;
return ;
}
void rotate(int u) {
int y = fa[u], z = fa[y], l = 0, r = 1;
if(z) ch[ch[1][z]==y][z] = u;
if(ch[1][y] == u) swap(l, r);
fa[u] = z; fa[y] = u; fa[ch[r][u]] = y;
ch[l][y] = ch[r][u]; ch[r][u] = y;
maintain(y); maintain(u);
return ;
}
void insert(int& o, int v) {
if(!o) {
ns[o = ++ToT] = Node(v, rand());
return maintain(o);
}
bool d = v > ns[o].v;
insert(ch[d][o], v); fa[ch[d][o]] = o;
if(ns[ch[d][o]].r > ns[o].r) {
int t = ch[d][o];
rotate(t); o = t;
}
return maintain(o);
}
void del(int& o, int v) {
if(!o) return ;
if(ns[o].v == v) {
if(!ch[0][o] && !ch[1][o]) o = 0;
else if(!ch[0][o]) {
int t = ch[1][o]; fa[t] = fa[o]; o = t;
}
else if(!ch[1][o]) {
int t = ch[0][o]; fa[t] = fa[o]; o = t;
}
else {
bool d = ns[ch[1][o]].r > ns[ch[0][o]].r;
int t = ch[d][o]; rotate(t); o = t;
del(ch[d^1][o], v);
}
}
else {
bool d = v > ns[o].v ;
del(ch[d][o], v);
}
return maintain(o);
}
int qrank(int o, int v) {
if(!o) return 0;
int ls = ch[0][o] ? ns[ch[0][o]].siz : 0;
if(v > ns[o].v) return ls + 1 + qrank(ch[1][o], v);
return qrank(ch[0][o], v);
}
#define err -233333333
#define errm 233333333
int qkth(int o, int k) {
if(!o) return err;
int ls = ch[0][o] ? ns[ch[0][o]].siz : 0;
if(k == ls + 1) return ns[o].v;
if(k > ls + 1) return qkth(ch[1][o], k - ls - 1);
return qkth(ch[0][o], k);
}
int qlow(int o, int v) {
if(!o) return err;
bool d = v > ns[o].v;
if(d) return max(ns[o].v, qlow(ch[d][o], v));
return qlow(ch[d][o], v);
}
int qupp(int o, int v) {
if(!o) return errm;
bool d = v >= ns[o].v;
if(!d) return min(ns[o].v, qupp(ch[d][o], v));
return qupp(ch[d][o], v);
}
int main() {
int q = read();
while(q--) {
int tp = read(), v = read();
if(tp == 1) insert(rt, v);
if(tp == 2) del(rt, v);
if(tp == 3) printf("%d
", qrank(rt, v) + 1);
if(tp == 4) printf("%d
", qkth(rt, v));
if(tp == 5) printf("%d
", qlow(rt, v));
if(tp == 6) printf("%d
", qupp(rt, v));
}
return 0;
}
再贴一个替罪羊树版本的。
#include <iostream>
#include <cstdio>
#include <algorithm>
#include <cmath>
#include <stack>
#include <vector>
#include <queue>
#include <cstring>
#include <string>
#include <map>
#include <set>
using namespace std;
const int BufferSize = 1 << 16;
char buffer[BufferSize], *Head, *Tail;
inline char Getchar() {
if(Head == Tail) {
int l = fread(buffer, 1, BufferSize, stdin);
Tail = (Head = buffer) + l;
}
return *Head++;
}
int read() {
int x = 0, f = 1; char c = Getchar();
while(!isdigit(c)){ if(c == '-') f = -1; c = Getchar(); }
while(isdigit(c)){ x = x * 10 + c - '0'; c = Getchar(); }
return x * f;
}
#define maxn 100010
#define oo 2147483647
struct Node {
int v, siz, reas, mx, mn;
bool del;
Node() {}
Node(int _): v(_), del(0) {}
} ns[maxn];
int rt, ToT, fa[maxn], ch[maxn][2];
void maintain(int o) {
ns[o].siz = ns[o].del ^ 1; ns[o].reas = 1;
ns[o].mx = ns[o].del ? -oo : ns[o].v;
ns[o].mn = ns[o].del ? oo : ns[o].v;
for(int i = 0; i < 2; i++) if(ch[o][i])
ns[o].siz += ns[ch[o][i]].siz,
ns[o].reas += ns[ch[o][i]].reas,
ns[o].mx = max(ns[o].mx, ns[ch[o][i]].mx),
ns[o].mn = min(ns[o].mn, ns[ch[o][i]].mn);
return ;
}
const double Bili = .6;
bool unbal(int o) {
return max(ch[o][0] ? ns[ch[o][0]].reas : 0, ch[o][1] ? ns[ch[o][1]].reas : 0) > Bili * ns[o].reas;
}
int rb;
void insert(int& o, int v) {
if(!o) {
ns[o = ++ToT] = Node(v);
return maintain(o);
}
bool d = v > ns[o].v;
insert(ch[o][d], v); fa[ch[o][d]] = o; maintain(o);
if(unbal(o)) rb = o;
return ;
}
int cntn, get[maxn];
void getnode(int o) {
if(!o) return ;
getnode(ch[o][0]);
if(!ns[o].del) get[++cntn] = o;
getnode(ch[o][1]);
fa[o] = ch[o][0] = ch[o][1] = 0;
return ;
}
void build(int& o, int l, int r) {
if(l > r){ o = 0; return ; }
int mid = l + r >> 1; o = get[mid];
build(ch[o][0], l, mid - 1); build(ch[o][1], mid + 1, r);
if(ch[o][0]) fa[ch[o][0]] = o;
if(ch[o][1]) fa[ch[o][1]] = o;
return maintain(o);
}
void rebuild(int& o) {
cntn = 0; getnode(o);
build(o, 1, cntn);
return ;
}
void Insert(int v) {
rb = 0; insert(rt, v);
if(!rb) return ;
int frb = fa[rb];
if(!frb) rebuild(rt), fa[rt] = 0;
else if(ch[frb][0] == rb) rebuild(ch[frb][0]), fa[ch[frb][0]] = frb;
else rebuild(ch[frb][1]), fa[ch[frb][1]] = frb;
return ;
}
bool unbal2(int o) {
return Bili * ns[o].reas > ns[o].siz;
}
void del(int o, int k) {
if(!o) return ;
int ls = ch[o][0] ? ns[ch[o][0]].siz : 0;
if(k == ls + 1 && !ns[o].del) {
ns[o].del = 1; maintain(o);
if(unbal2(o)) rb = o;
return ;
}
if(k > ls + (ns[o].del ^ 1)) {
del(ch[o][1], k - ls - (ns[o].del ^ 1)); maintain(o);
if(unbal2(o)) rb = o;
return ;
}
del(ch[o][0], k); maintain(o);
if(unbal2(o)) rb = o;
return ;
}
int Find(int o, int v) {
if(!o) return 0;
int ls = ch[o][0] ? ns[ch[o][0]].siz : 0;
if(v <= ns[o].v) return Find(ch[o][0], v);
return ls + (ns[o].del ^ 1) + Find(ch[o][1], v);
}
void Delete(int v) {
int id = Find(rt, v) + 1;
rb = 0; del(rt, id);
int tmp = ns[rt].reas;
if(!rb) return ;
int frb = fa[rb];
if(!frb) rebuild(rt), fa[rt] = 0;
else if(ch[frb][0] == rb) rebuild(ch[frb][0]), fa[ch[frb][0]] = frb;
else rebuild(ch[frb][1]), fa[ch[frb][1]] = frb;
return ;
}
int qkth(int o, int k) {
if(!o) return 0;
int ls = ch[o][0] ? ns[ch[o][0]].siz : 0;
if(k == ls + 1 && !ns[o].del) return ns[o].v;
if(k > ls + (ns[o].del ^ 1)) return qkth(ch[o][1], k - ls - (ns[o].del ^ 1));
return qkth(ch[o][0], k);
}
int qlow(int o, int v) {
if(!o) return -oo;
if(ns[o].v < v) return max(max(ns[o].del ? -oo : ns[o].v, ch[o][0] ? ns[ch[o][0]].mx : -oo), qlow(ch[o][1], v));
return qlow(ch[o][0], v);
}
int qupp(int o, int v) {
if(!o) return oo;
if(ns[o].v > v) return min(min(ns[o].del ? oo : ns[o].v, ch[o][1] ? ns[ch[o][1]].mn : oo), qupp(ch[o][0], v));
return qupp(ch[o][1], v);
}
int main() {
int q = read();
while(q--) {
int tp = read(), x = read();
if(tp == 1) Insert(x);
if(tp == 2) Delete(x);
if(tp == 3) printf("%d
", Find(rt, x) + 1);
if(tp == 4) printf("%d
", qkth(rt, x));
if(tp == 5) printf("%d
", qlow(rt, x));
if(tp == 6) printf("%d
", qupp(rt, x));
}
return 0;
}
懒惰删除真的一点都不懒惰！！！打了删除标记反而更难处理了。。。太多细节要考虑。
再贴一个 fhq treap 的版本。
#include <iostream>
#include <cstdio>
#include <cstdlib>
#include <cstring>
#include <cctype>
#include <algorithm>
#include <cassert>
using namespace std;
#define rep(i, s, t) for(int i = (s), mi = (t); i <= mi; i++)
#define dwn(i, s, t) for(int i = (s), mi = (t); i >= mi; i--)
const int BufferSize = 1 << 16;
char buffer[BufferSize], *Head, *Tail;
inline char Getchar() {
if(Head == Tail) {
int l = fread(buffer, 1, BufferSize, stdin);
Tail = (Head = buffer) + l;
}
return *Head++;
}
int read() {
int x = 0, f = 1; char c = Getchar();
while(!isdigit(c)){ if(c == '-') f = -1; c = Getchar(); }
while(isdigit(c)){ x = x * 10 + c - '0'; c = Getchar(); }
return x * f;
}
#define maxn 100010
#define oo 2147483647
#define pii pair <int, int>
#define x first
#define y second
#define mp(x, y) make_pair(x, y)
struct Node {
int v, r, siz;
Node() {}
Node(int _v): v(_v), r(rand()), siz(1) {}
} ns[maxn];
int ToT, rt, ch[maxn][2];
void maintain(int o) {
if(!o) return ;
ns[o].siz = 1;
if(ch[o][0]) ns[o].siz += ns[ch[o][0]].siz;
if(ch[o][1]) ns[o].siz += ns[ch[o][1]].siz;
return ;
}
int merge(int a, int b) { // max{a} <= min{b}
if(!a) return maintain(b), b;
if(!b) return maintain(a), a;
if(ns[a].r > ns[b].r) return ch[a][1] = merge(ch[a][1], b), maintain(a), a;
return ch[b][0] = merge(a, ch[b][0]), maintain(b), b;
}
pii split(int o, int k) {
if(!o) return mp(0, 0);
if(!k) return mp(0, o);
int ls = ch[o][0] ? ns[ch[o][0]].siz : 0;
if(k <= ls) {
pii pr = split(ch[o][0], k);
ch[o][0] = pr.y; maintain(o);
return mp(pr.x, o);
}
pii pr = split(ch[o][1], k - ls - 1);
ch[o][1] = pr.x; maintain(o);
return mp(o, pr.y);
}
int Find(int o, int v) {
if(!o) return 0;
int ls = ch[o][0] ? ns[ch[o][0]].siz : 0;
if(v <= ns[o].v) return Find(ch[o][0], v);
return ls + 1 + Find(ch[o][1], v);
}
void Insert(int v) {
int rnk = Find(rt, v);
pii pr = split(rt, rnk);
ns[++ToT] = Node(v);
rt = merge(pr.x, ToT);
rt = merge(rt, pr.y);
return ;
}
void Delete(int& o, int v) {
if(!o) return ;
if(ns[o].v == v) return (void)(o = merge(ch[o][0], ch[o][1]));
if(v < ns[o].v) Delete(ch[o][0], v);
else Delete(ch[o][1], v);
return maintain(o);
}
int qkth(int o, int k) {
assert(o);
int ls = ch[o][0] ? ns[ch[o][0]].siz : 0;
if(k == ls + 1) return ns[o].v;
if(k <= ls) return qkth(ch[o][0], k);
return qkth(ch[o][1], k - ls - 1);
}
int qpre(int o, int v) {
if(!o) return -oo;
if(ns[o].v < v) return max(ns[o].v, qpre(ch[o][1], v));
return qpre(ch[o][0], v);
}
int qnxt(int o, int v) {
if(!o) return oo;
if(ns[o].v > v) return min(ns[o].v, qnxt(ch[o][0], v));
return qnxt(ch[o][1], v);
}
int main () {
srand(5);
int q = read();
while(q--) {
int tp = read();
if(tp == 1) Insert(read());
if(tp == 2) Delete(rt, read());
if(tp == 3) printf("%d
", Find(rt, read()) + 1);
if(tp == 4) printf("%d
", qkth(rt, read()));
if(tp == 5) printf("%d
", qpre(rt, read()));
if(tp == 6) printf("%d
", qnxt(rt, read()));
}
return 0;
}