Splay树简单操作

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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

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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