前几天刚刚自学了一下splay，发现思路真简单实现起来好麻烦
先贴一下头文件
# include <stdio.h>
# include <stdlib.h>
# include <iostream>
# include <string.h>
# define ll long long
# define RG register//卡常
# define IL inline//再卡常
# define UN unsigned
# define mem(a, b) memset(a, b, sizeof(a))
# define min(a, b) ((a) < (b)) ? (a) : (b)
# define max(a, b) ((a) > (b)) ? (a) : (b)
using namespace std;-
核心旋转操作 Splay操作（之后的操作基本上都要用到）
1.当旋转的节点为爷爷节点的左儿子的左儿子时
进行两次右旋操作，先转父亲，再转自己
2.当旋转的节点为爷爷节点的左儿子的右儿子时
进行一次左旋操作，一次右旋操作，都转自己
3.如果要转到的点的为爷爷节点的右儿子，直接左旋
4.剩下两种情况把以上两种情况左右互换即可
贴一段代码
IL void Rot(RG tree *x, RG int i){ //0为左旋，1为右旋，0为左儿子，1为右儿子
RG tree *y = x -> fa;
y -> ch[!i] = x -> ch[i];
if(x -> ch[i] != NULL) x -> ch[i] -> fa = y;
x -> fa = y -> fa;
if(y -> fa != NULL)
if(y -> fa -> ch[0] == y) y -> fa -> ch[0] = x;
else y -> fa -> ch[1] = x;
x -> ch[i] = y; y -> fa = x;
if(y == root) root = x;
}
IL void Splay(RG tree *x, RG tree *f){
while(x -> fa != f)
if(x -> fa -> fa == f)
if(x == x -> fa -> ch[0]) Rot(x, 1);
else Rot(x, 0);
else{
RG tree *y = x -> fa, *z = y -> fa;
if(z -> ch[0] == y)
if(y -> ch[0] == x) Rot(y, 1), Rot(x, 1);
else Rot(x, 0), Rot(x, 1);
else
if(y -> ch[1] == x) Rot(y, 0), Rot(x, 0);
else Rot(x, 1), Rot(x, 0);
}
}- 插入操作
和二叉排序树一样，只不过弄完后把它Splay到根（不要问为什么）
丑陋的代码
IL void Insert(RG int num){
RG tree *x = root;
if(x == NULL){
x = new tree;
x -> val = num;
root = x;
}
else while(2333)
if(num < x -> val){
if(x -> ch[0] == NULL){
x -> ch[0] = new tree;
x -> ch[0] -> fa = x;
x = x -> ch[0];
x -> val = num;
break;
}
x = x -> ch[0];
}
else if(num > x -> val){
if(x -> ch[1] == NULL){
x -> ch[1] = new tree;
x -> ch[1] -> fa = x;
x = x -> ch[1];
x -> val = num;
break;
}
x = x -> ch[1];
}
Splay(x, NULL);
}- 查找
与二叉排序树一样
IL void Find(RG int num){
RG tree *x = root;
while(x -> val != num)
if(num < x -> val) x = x -> ch[0];
else x = x -> ch[1];
Splay(x, NULL);
}- 查找前驱和后缀
前驱，跳到它的左儿子再不停地跳右儿子
后继，跳到它的右儿子再不停地跳左儿子
IL void Findmx(RG tree *x, RG tree *f, RG int i){ //0表示后继，1表示前驱，f为该节点，x为它的左或右儿子1
while(x -> ch[i] != NULL) x = x -> ch[i];
Splay(x, f);
}
- 删除操作
先找到数字的位置，Splay到根，删掉它，找它左子树中的最大数（前驱）Splay到它下面作为新的根，连接右子树即可
代码
IL void Delete(RG int num){
Find(num);
RG tree *x = root;
else if(x -> ch[0] == NULL || x -> ch[1] == NULL)
if(x -> ch[0] != NULL) root = x -> ch[0], root -> fa = NULL;
else if(x -> ch[1] != NULL) root = x -> ch[1], root -> fa = NULL;
else root = NULL;
else{
Findmx(x -> ch[0], x, 1);
root = x -> ch[0];
root -> fa = NULL;
root -> ch[1] = x -> ch[1];
if(root -> ch[1] != NULL) root -> ch[1] -> fa = root;
}
}- 查找某数的排名
实行查找操作，排名就是他左子树大小加一
IL int Size(RG tree *x){
return (x == NULL) ? 0 : x -> size + 1;
}
IL int Rank(RG int num){
Find(num);
return Size(root -> ch[0]) + 1;
}- 查找排名为k的数
若大于当前的左子树大小加一，跳右儿子，k -= 左子树大小加一；
否则跳右节点
IL int Pos(RG int num){
RG tree *x = root;
while(2333){
RG int l = Size(x -> ch[0]);
if(num == l + 1) break;
if(num <= l) x = x -> ch[0];
else{
num -= (l + 1);
x = x -> ch[1];
}
}
return x -> val;
}以上就是基本操作
- 关于更新
如子树大小
IL int Size(RG tree *x){
return (x == NULL) ? 0 : x -> size + 1;
}
IL void Updata(RG tree *x){
if(x == NULL) return;
x -> size = Size(x -> ch[0]) + Size(x -> ch[1]);
}
IL void Rot(RG tree *x, RG int i){ //0为左旋，1为右旋
RG tree *y = x -> fa;
y -> ch[!i] = x -> ch[i];
if(x -> ch[i] != NULL) x -> ch[i] -> fa = y;
x -> fa = y -> fa;
if(y -> fa != NULL)
if(y -> fa -> ch[0] == y) y -> fa -> ch[0] = x;
else y -> fa -> ch[1] = x;
x -> ch[i] = y; y -> fa = x;
Updata(y); //大佬说写在这里
if(y == root) root = x;
}
IL void Splay(RG tree *x, RG tree *f){
while(x -> fa != f)
if(x -> fa -> fa == f)
if(x == x -> fa -> ch[0]) Rot(x, 1);
else Rot(x, 0);
else{
RG tree *y = x -> fa, *z = y -> fa;
if(z -> ch[0] == y)
if(y -> ch[0] == x) Rot(y, 1), Rot(x, 1);
else Rot(x, 0), Rot(x, 1);
else
if(y -> ch[1] == x) Rot(y, 0), Rot(x, 0);
else Rot(x, 1), Rot(x, 0);
}
Updata(x); //大佬说要写在这里
}
简单的栗子：
链接bzoj3224
请读者思考2分钟
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好了直接看代码 代码调了好久QAQ
# include <stdio.h>
# include <stdlib.h>
# include <iostream>
# include <string.h>
# define ll long long
# define RG register
# define IL inline
# define UN unsigned
# define mem(a, b) memset(a, b, sizeof(a))
# define min(a, b) ((a) < (b)) ? (a) : (b)
# define max(a, b) ((a) > (b)) ? (a) : (b)
using namespace std;
IL int Get(){
char c = '!'; int z = 1, num = 0;
while(c != '-' && (c < '0' || c > '9'))
c = getchar();
if(c == '-')
z = -1, c = getchar();
while(c >= '0' && c <= '9')
num = num * 10 + c - '0', c = getchar();
return num * z;
}
struct tree{
tree *fa, *ch[2];
int val, tot, size;//tot用来判断重复的数
IL tree(){
fa = ch[0] = ch[1] = NULL;
size = val = tot = 0;
}
} *root;
IL int Size(RG tree *x){
return (x == NULL) ? 0 : x -> size + x -> tot + 1;
}
IL void Updata(RG tree *x){
if(x == NULL) return;
x -> size = Size(x -> ch[0]) + Size(x -> ch[1]);
}
IL void Rot(RG tree *x, RG int i){ //0为左旋，1为右旋
RG tree *y = x -> fa;
y -> ch[!i] = x -> ch[i];
if(x -> ch[i] != NULL) x -> ch[i] -> fa = y;
x -> fa = y -> fa;
if(y -> fa != NULL)
if(y -> fa -> ch[0] == y) y -> fa -> ch[0] = x;
else y -> fa -> ch[1] = x;
x -> ch[i] = y; y -> fa = x; Updata(y);
if(y == root) root = x;
}
IL void Splay(RG tree *x, RG tree *f){
while(x -> fa != f)
if(x -> fa -> fa == f)
if(x == x -> fa -> ch[0]) Rot(x, 1);
else Rot(x, 0);
else{
RG tree *y = x -> fa, *z = y -> fa;
if(z -> ch[0] == y)
if(y -> ch[0] == x) Rot(y, 1), Rot(x, 1);
else Rot(x, 0), Rot(x, 1);
else
if(y -> ch[1] == x) Rot(y, 0), Rot(x, 0);
else Rot(x, 1), Rot(x, 0);
}
Updata(x);
}
IL void Insert(RG int num){
RG tree *x = root;
if(x == NULL){
x = new tree;
x -> val = num;
root = x;
}
else if(num == x -> val) x -> tot++, root = x;
else while(2333)
if(num < x -> val){
if(x -> ch[0] == NULL){
x -> ch[0] = new tree;
x -> ch[0] -> fa = x;
x = x -> ch[0];
x -> val = num;
break;
}
x = x -> ch[0];
}
else if(num > x -> val){
if(x -> ch[1] == NULL){
x -> ch[1] = new tree;
x -> ch[1] -> fa = x;
x = x -> ch[1];
x -> val = num;
break;
}
x = x -> ch[1];
}
else{
x -> tot++;
break;
}
Splay(x, NULL);
}
IL void Find(RG int num){
RG tree *x = root;
while(x -> val != num)
if(num < x -> val) x = x -> ch[0];
else x = x -> ch[1];
Splay(x, NULL);
}
IL void Findmx(RG tree *x, RG tree *f, RG int i){
while(x -> ch[i] != NULL) x = x -> ch[i];
Splay(x, f);
}
IL void Delete(RG int num){
Find(num);
RG tree *x = root;
if(root -> tot) root -> tot--;
else if(x -> ch[0] == NULL || x -> ch[1] == NULL)
if(x -> ch[0] != NULL) root = x -> ch[0], root -> fa = NULL;
else if(x -> ch[1] != NULL) root = x -> ch[1], root -> fa = NULL;
else root = NULL;
else{
Findmx(x -> ch[0], x, 1);
root = x -> ch[0];
root -> fa = NULL;
root -> ch[1] = x -> ch[1];
if(root -> ch[1] != NULL) root -> ch[1] -> fa = root;
}
}
IL int Rank(RG int num){
Find(num);
return Size(root -> ch[0]) + 1;
}
IL int Pos(RG int num){
RG tree *x = root;
while(2333){
RG int l = Size(x -> ch[0]);
if(num > l && num <= l + x -> tot + 1) break;
if(num <= l) x = x -> ch[0];
else{
num -= (l + x -> tot + 1);
x = x -> ch[1];
}
}
return x -> val;
}
int main(){
RG int n = Get(), opt, x;
while(n--){
opt = Get(); x = Get();
if(opt == 1) Insert(x);
if(opt == 2) Delete(x);
if(opt == 3) printf("%d\n", Rank(x));
if(opt == 4) printf("%d\n", Pos(x));
//找前驱后缀：插入数后再删除，显然有更快的（不想打其他方法了，反正能过）
if(opt == 5){
Insert(x);
Findmx(root -> ch[0], NULL, 1);
printf("%d\n", root -> val);
Delete(x);
}
if(opt == 6){
Insert(x);
Findmx(root -> ch[1], NULL, 0);
printf("%d\n", root -> val);
Delete(x);
}
}
return 0;
}关于区间操作
一张丑陋的图
把l-1splay到根，r+1splay到根的右儿子，则图中那个丑陋的子树就是要求的[l，r]了。删除区间
直接断开边（显然浪费空间，自己想办法目前没遇到MLE的情况）- 翻转区间
用类似于线段树的懒懒的lazy标记，每次Find，splay等操作时把标记下放，更换两个子树
又一个栗子
链接bzoj3223
再思考两分钟
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看代码
# include <stdio.h>
# include <stdlib.h>
# include <iostream>
# include <string.h>
# define ll long long
# define RG register
# define IL inline
# define UN unsigned
# define mem(a, b) memset(a, b, sizeof(a))
# define min(a, b) ((a) < (b)) ? (a) : (b)
# define max(a, b) ((a) > (b)) ? (a) : (b)
using namespace std;
IL int Get(){
char c = '!'; int z = 1, num = 0;
while(c != '-' && (c < '0' || c > '9'))
c = getchar();
if(c == '-')
z = -1, c = getchar();
while(c >= '0' && c <= '9')
num = num * 10 + c - '0', c = getchar();
return num * z;
}
struct tree{
tree *fa, *ch[2];
int size, lazy, pos;
IL tree(){
fa = ch[0] = ch[1] = NULL;
pos = lazy = size = 0;
}
IL void Pushdown(){//下放
if(!lazy) return;
lazy = 0;
if(ch[1] == NULL && ch[0] == NULL) return;
if(ch[0] != NULL) ch[0] -> lazy ^= 1;
if(ch[1] != NULL) ch[1] -> lazy ^= 1;
swap(ch[0], ch[1]);
}
} *root;
int n;
IL int Size(RG tree *x){
return (x == NULL) ? 0 : x -> size + 1;
}
IL void Updata(RG tree *x){
if(x == NULL) return;
x -> size = Size(x -> ch[0]) + Size(x -> ch[1]);
}
IL void Dfs(RG tree *x){
if(x == NULL) return;
x -> Pushdown();
Dfs(x -> ch[0]);
if(x -> pos && x -> pos <= n)
printf("%d ", x -> pos);
Dfs(x -> ch[1]);
}
IL void Rot(RG tree *x, RG int i){ //0为左旋，1为右旋
RG tree *y = x -> fa;
x -> Pushdown(); y -> Pushdown();
y -> ch[!i] = x -> ch[i];
if(x -> ch[i] != NULL) x -> ch[i] -> fa = y;
x -> fa = y -> fa;
if(y -> fa != NULL)
if(y -> fa -> ch[0] == y) y -> fa -> ch[0] = x;
else y -> fa -> ch[1] = x;
x -> ch[i] = y; y -> fa = x; Updata(y);
if(y == root) root = x;
}
IL void Splay(RG tree *x, RG tree *f){
while(x -> fa != f){
x -> Pushdown();
if(x -> fa -> fa == f)
if(x == x -> fa -> ch[0]) Rot(x, 1);
else Rot(x, 0);
else{
RG tree *y = x -> fa, *z = y -> fa;
if(z -> ch[0] == y)
if(y -> ch[0] == x) Rot(y, 1), Rot(x, 1);
else Rot(x, 0), Rot(x, 1);
else
if(y -> ch[1] == x) Rot(y, 0), Rot(x, 0);
else Rot(x, 1), Rot(x, 0);
}
}
Updata(x);
}
IL tree *Build(RG int l, RG int r, RG tree *f){
RG int mid = (l + r) >> 1;
tree *x = new tree;
x -> pos = mid;
x -> fa = f;
if(mid > l) x -> ch[0] = Build(l, mid - 1, x);
if(mid < r) x -> ch[1] = Build(mid + 1, r, x);
Updata(x);
return x;
}
IL void Find(RG int num, RG tree *f){
RG tree *x = root;
while(2333){
x -> Pushdown();
RG int l = Size(x -> ch[0]);
if(num < l) x = x -> ch[0];
else if(num == l) break;
else{
num -= (l + 1);
x = x -> ch[1];
}
}
Splay(x, f);
}
IL void Turn(){
RG int l = Get(), r = Get();
Find(l - 1, NULL); Find(r + 1, root);
root -> ch[1] -> ch[0] -> lazy ^= 1;
}
int main(){
n = Get();
RG int m = Get();
root = Build(0, n + 1, NULL);
while(m--) Turn();
Dfs(root);
printf("\n");
return 0;
}解释或代码错误还请大佬指出，本蒟蒻一定会改