Splay树(区间添加删除 | 区间翻转)——HDU 3487 Play with Chain

2024-09-05 04:08

本文主要是介绍Splay树(区间添加删除 | 区间翻转)——HDU 3487 Play with Chain,希望对大家解决编程问题提供一定的参考价值,需要的开发者们随着小编来一起学习吧!

对应HDU题目:点击打开链接

Play with Chain

Time Limit: 6000/2000 MS (Java/Others)    Memory Limit: 65536/32768 K (Java/Others)
Total Submission(s): 4571    Accepted Submission(s): 1859


Problem Description
YaoYao is fond of playing his chains. He has a chain containing n diamonds on it. Diamonds are numbered from 1 to n.
At first, the diamonds on the chain is a sequence: 1, 2, 3, …, n.
He will perform two types of operations:
CUT a b c: He will first cut down the chain from the ath diamond to the bth diamond. And then insert it after the cth diamond on the remaining chain.
For example, if n=8, the chain is: 1 2 3 4 5 6 7 8; We perform “CUT 3 5 4”, Then we first cut down 3 4 5, and the remaining chain would be: 1 2 6 7 8. Then we insert “3 4 5” into the chain before 5th diamond, the chain turns out to be: 1 2 6 7 3 4 5 8.

FLIP a b: We first cut down the chain from the ath diamond to the bth diamond. Then reverse the chain and put them back to the original position.
For example, if we perform “FLIP 2 6” on the chain: 1 2 6 7 3 4 5 8. The chain will turn out to be: 1 4 3 7 6 2 5 8

He wants to know what the chain looks like after perform m operations. Could you help him? 

Input
There will be multiple test cases in a test data. 
For each test case, the first line contains two numbers: n and m (1≤n, m≤3*100000), indicating the total number of diamonds on the chain and the number of operations respectively.
Then m lines follow, each line contains one operation. The command is like this:
CUT a b c // Means a CUT operation, 1 ≤ a ≤ b ≤ n, 0≤ c ≤ n-(b-a+1).
FLIP a b    // Means a FLIP operation, 1 ≤ a < b ≤ n.
The input ends up with two negative numbers, which should not be processed as a case.

Output
For each test case, you should print a line with n numbers. The ith number is the number of the ith diamond on the chain.

Sample Input
  
8 2 CUT 3 5 4 FLIP 2 6 -1 -1

Sample Output
  
1 4 3 7 6 2 5 8

题意:

n个数一开始是1~n顺序排列,有两个操作。

CUT(a, b, c)操作表示把第a到第b个数取下放到新组成的第c个数后面;

FLIP(a, b)操作表示把第a到第b个数翻转;

比如样例: n = 8

1   2   3   4   5   6   7   8  

CUT(3,  5,  4)后数列变成 1   2   6   7   3   4   5   8

FLIP(2,  6)后数列变成1   4   3   7   6   2   5   8

问m次操作后的数列为?


思路:

伸展树的基础操作,区间截断,区间翻转。

区间截断对于CUT(a, b, c)

1)提取区间[a, b]

具体方法:Splay(a - 1, T),Splay(b +1, T->right);即把第a - 1个数旋转到根,把第b + 1个数旋转到根的右儿子;那以根的右儿子的左儿子为根的子树就是所有区间[a, b]内的值;把它剪下。

2)把第c个数旋转到根,把第c + 1个数旋转到根的右儿子;那根的右儿子的左儿子肯定是空的。

3)把剪下的子树接在根的右儿子的左儿子。

区间翻转:对于FLIP(a, b)

1)提取区间[a, b]

2)在左儿子做翻转标志(注意不是简单的赋值为1,而是要做异或操作,即原来是1的标记为0,是0的标记为1)

3)在适当的地方加Push_down向下更新(跟线段树区间更新一样)


首先要明白的一点是二叉树结点的信息是最终中序遍历(一定是按下标先后顺序)输出的值,而不是下标的值;那怎样取下标为a的结点呢?

利用每个结点的sz,它表示以该结点为子树的的所有元素个数(T->sz = T->left->sz + T->right->sz + 1)

从根结点p开始

1)如果(左子树的元素个数加上1(1表示p结点) )sum = p->left->sz + 1;if (sum == a) ,那就找到了,返回p结点

2)sum比a小,说明要往右走(p = p->right);同时a -= sum;跳到步骤1

3)sum比a大,说明要往左走(p = p->right);a无需变动;跳到步骤1


怎样书写Push_down函数

就把左右子树对调,然后取消该结点标记,把两棵子树添加标记。


注意:修改时要注意Push_down的地方,注意父亲指针;注意更新sz;用动态的方法时要注意儿子是否为空。


指针型

#include <stdio.h>
#include <stdlib.h>
#include <string.h>
#include <math.h>
#define MAX(x, y) ((x) > (y) ? (x) : (y))typedef struct TREE
{int data;TREE *fa, *l, *r;int sz; //以该结点为根的树的总结点数bool flag; //翻转标记
}Tree;bool space;void Push_down(Tree *T)
{if(NULL == T) return;//左右子树对调if(T->flag){Tree *tmp;tmp = T->r;T->r = T->l;T->l = tmp;T->flag = 0;if(T->l) T->l->flag ^= 1;if(T->r) T->r->flag ^= 1;}
}void Init(Tree *&T, int n)
{int i;bool k = 0;Tree *cur, *pre;for(i = n + 1; i > -1; i--){cur = (Tree *)malloc(sizeof(Tree));cur->data = i;cur->fa = cur->l = cur->r = NULL;cur->sz = 1;cur->flag = 0;if(k){cur->r = pre;pre->fa = cur;cur->sz = pre->sz + 1;}if(!k) k = 1;pre = cur;}T = cur;
}void PreOrder(Tree *T)
{if(NULL == T) return;printf("%d ", T->data);PreOrder(T->l);PreOrder(T->r);
}void MidOrder(Tree *T, int n)
{if(NULL == T) return;Push_down(T);MidOrder(T->l, n);if(T->data > 0 && T->data < n + 1){if(space) printf(" ");else space = 1;printf("%d", T->data);}MidOrder(T->r, n);
}void R_rotate(Tree *x)
{Tree *y = x->fa;Tree *z = y->fa;Tree *k = x->r;int sx = x->sz, sy = y->sz, sk = 0;if(k) sk = k->sz;y->l = k;x->r = y;if(z){if(y == z->l) z->l = x;else z->r = x;}if(k) k->fa = y;y->fa = x;x->fa = z;y->sz = sy - sx + sk;x->sz = sx - sk + y->sz;
}void L_rotate(Tree *x)
{Tree *y = x->fa;Tree *z = y->fa;Tree *k = x->l;int sx = x->sz, sy = y->sz, sk = 0;if(k) sk = k->sz;y->r = k;x->l = y;if(z){if(y == z->r) z->r = x;else z->l = x;}if(k) k->fa = y;y->fa = x;x->fa = z;y->sz = sy - sx + sk;x->sz = sx - sk + y->sz;
}//寻找第x个数的结点
Tree *FindTag(Tree *T, int x)
{if(NULL == T) return NULL;Push_down(T);Tree *p;p = T;int sum = (p->l ? p->l->sz : 0) + 1;while(sum != x && p){if(sum < x){p = p->r;x -= sum;}else p = p->l;Push_down(p);sum = (p->l ? p->l->sz : 0) + 1;}Push_down(p);return p;
}void Splay(int x, Tree *&T)
{Push_down(T);Tree *p, *X, *end, *new_t;end = T->fa;new_t = T;if(end) new_t = T->fa;X = FindTag(new_t, x);while(X->fa != end){p = X->fa;if(end == p->fa){ //p是根结点if(X == p->l) R_rotate(X);else L_rotate(X);break;}//p不是根结点if(X == p->l){if(p == p->fa->l){R_rotate(p); //LLR_rotate(X); //LL}else{R_rotate(X); //RLL_rotate(X);}}else{if(p == p->fa->r){ //RRL_rotate(p);L_rotate(X);}else{ //LRL_rotate(X);R_rotate(X);}}}T = X;
}void CUT(Tree *&T, int a, int b, int c)
{//取[a,b]Splay(a - 1, T);Splay(b + 1, T->r);//剪[a,b]Tree *tmp;tmp = T->r->l;tmp->fa = NULL;T->r->l = NULL;T->r->sz -= tmp->sz;T->sz -= tmp->sz;//移动第c个数到根结点,第c+1个数到根结点右儿子//这样根结点右儿子的左儿子必然为空,就可以把剪掉的放上去Splay(c, T);Splay(c + 1, T->r);//接[a, b]T->r->l = tmp;tmp->fa = T->r;T->r->sz += tmp->sz;T->sz += tmp->sz;
}void FLIP(Tree *&T, int a, int b)
{//取[a,b]Splay(a - 1, T);Splay(b + 1, T->r);//标记T->r->lT->r->l->flag ^= 1;
}void FreeTree(Tree *T)
{if(NULL == T) return;FreeTree(T->l);FreeTree(T->r);free(T);
}int main()
{//freopen("in.txt", "r", stdin);Tree *T;int n, q, a, b, c;char s[6];while(scanf("%d%d", &n, &q), n >= 0 && q >= 0){space = 0;T = NULL;Init(T, n);while(q--){scanf("%s", s);if('C' == s[0]){scanf("%d%d%d", &a, &b, &c);CUT(T, a + 1, b + 1, c + 1);}else{scanf("%d%d", &a, &b);FLIP(T, a + 1, b + 1);}}MidOrder(T, n);printf("\n");FreeTree(T);}return 0;
}

数组型

#include <stdio.h>
#include <stdlib.h>
#include <string.h>
#include <math.h>
#define MAX(x, y) ((x) > (y) ? (x) : (y))
#define M 300090
#define nul (1<<30)
int data[M];
int Left[M];
int Right[M];
int fa[M];
int sz[M];
bool flag[M];
bool space;void Init()
{int i;for(i = 0; i < M; i++)Left[i] = Right[i] = fa[i] = nul;
}void Push_down(int T)
{if(nul == T) return;if(flag[T]){int tmp = Right[T];Right[T] = Left[T];Left[T] = tmp;flag[T] = 0;if(nul != Left[T]) flag[Left[T]] ^= 1;if(nul != Right[T]) flag[Right[T]] ^= 1;}
}void Create(int &T, int n)
{bool k = 0;int cur, pre;for(cur = n + 1; cur > -1; cur--){data[cur] = cur;sz[cur] = 1;flag[cur] = 0;if(k){Right[cur] = pre;fa[pre] = cur;sz[cur] = sz[pre] + 1;}if(!k) k = 1;pre = cur;}T = pre;
}void InOrder(int T, int n)
{if(nul == T) return;Push_down(T);InOrder(Left[T], n);if(data[T] > 0 && data[T] < n + 1){if(space) printf(" ");else space = 1;printf("%d", data[T]);}InOrder(Right[T], n);
}void PreOrder(int T)
{if(nul == T) return;printf("%d ", data[T]);PreOrder(Left[T]);PreOrder(Right[T]);
}void R_rotate(const int x)
{const int y = fa[x];const int z = fa[y];const int k = Right[x];int sx = sz[x], sy = sz[y], sk = 0;if(nul != k) sk = sz[k];Left[y] = k;Right[x] = y;if(nul != z){if(y == Left[z]) Left[z] = x;else Right[z] = x;}if(nul != k) fa[k] = y;fa[y] = x;fa[x] = z;sz[y] = sy - sx + sk;sz[x] = sx - sk + sz[y];
}void L_rotate(const int x)
{const int y = fa[x];const int z = fa[y];const int k = Left[x];int sx = sz[x], sy = sz[y], sk = 0;if(nul != k) sk = sz[k];Right[y] = k;Left[x] = y;if(nul != z){if(y == Right[z]) Right[z] = x;else Left[z] = x;}if(nul != k) fa[k] = y;fa[y] = x;fa[x] = z;sz[y] = sy - sx + sk;sz[x] = sx - sk + sz[y];
}int FindTag(int T, int x)
{if(nul == T) return nul;Push_down(T);int p = T;int sum = (nul != Left[p] ? sz[Left[p]] : 0) + 1;while(sum != x && nul != p){if(sum < x){p = Right[p];x -= sum;}else p = Left[p];Push_down(p);sum = (nul != Left[p] ? sz[Left[p]] : 0) + 1;}Push_down(p);return p;
}void Splay(int x, int &T)
{Push_down(T);int p, end, new_t;end = fa[T];new_t = T;if(nul != end) new_t = fa[T];x = FindTag(new_t, x);while(end != fa[x]){p = fa[x];if(end == fa[p]){ //p是根结点if(x == Left[p]) R_rotate(x);else L_rotate(x);break;}//p不是根结点if(x == Left[p]){if(p == Left[fa[p]]){R_rotate(p); //LLR_rotate(x); //LL}else{R_rotate(x); //RLL_rotate(x);}}else{if(p == Right[fa[p]]){ //RRL_rotate(p);L_rotate(x);}else{ //LRL_rotate(x);R_rotate(x);}}}T = x;
}void CUT(int &T, int a, int b, int c)
{Splay(a - 1, T);Splay(b + 1, Right[T]);int tmp;tmp = Left[Right[T]];fa[tmp] = Left[Right[T]] = nul;sz[Right[T]] -= sz[tmp];sz[T] -= sz[tmp];Splay(c, T);Splay(c + 1, Right[T]);Left[Right[T]] = tmp;fa[tmp] = Right[T];sz[Right[T]] += sz[tmp];sz[T] += sz[tmp];
}void FLIP(int &T, int a, int b)
{Splay(a - 1, T);Splay(b + 1, Right[T]);flag[Left[Right[T]]] ^= 1;
}int main()
{//freopen("in.txt", "r", stdin);int T;int n, q, a, b, c;char s[6];while(scanf("%d%d", &n, &q), n >= 0 && q >= 0){space = 0;T = nul;Init();Create(T, n);while(q--){scanf("%s", s);if('C' == s[0]){scanf("%d%d%d", &a, &b, &c);CUT(T, a + 1, b + 1, c + 1);}else{scanf("%d%d", &a, &b);FLIP(T, a + 1, b + 1);}}InOrder(T, n);printf("\n");}return 0;
}







这篇关于Splay树(区间添加删除 | 区间翻转)——HDU 3487 Play with Chain的文章就介绍到这儿,希望我们推荐的文章对编程师们有所帮助!



http://www.chinasem.cn/article/1137905

相关文章

macOS无效Launchpad图标轻松删除的4 种实用方法

《macOS无效Launchpad图标轻松删除的4种实用方法》mac中不在appstore上下载的应用经常在删除后它的图标还残留在launchpad中,并且长按图标也不会出现删除符号,下面解决这个问... 在 MACOS 上,Launchpad(也就是「启动台」)是一个便捷的 App 启动工具。但有时候,应

Mysql删除几亿条数据表中的部分数据的方法实现

《Mysql删除几亿条数据表中的部分数据的方法实现》在MySQL中删除一个大表中的数据时,需要特别注意操作的性能和对系统的影响,本文主要介绍了Mysql删除几亿条数据表中的部分数据的方法实现,具有一定... 目录1、需求2、方案1. 使用 DELETE 语句分批删除2. 使用 INPLACE ALTER T

C++从序列容器中删除元素的四种方法

《C++从序列容器中删除元素的四种方法》删除元素的方法在序列容器和关联容器之间是非常不同的,在序列容器中,vector和string是最常用的,但这里也会介绍deque和list以供全面了解,尽管在一... 目录一、简介二、移除给定位置的元素三、移除与某个值相等的元素3.1、序列容器vector、deque

C++原地删除有序数组重复项的N种方法

《C++原地删除有序数组重复项的N种方法》给定一个排序数组,你需要在原地删除重复出现的元素,使得每个元素只出现一次,返回移除后数组的新长度,不要使用额外的数组空间,你必须在原地修改输入数组并在使用O(... 目录一、问题二、问题分析三、算法实现四、问题变体:最多保留两次五、分析和代码实现5.1、问题分析5.

SQL Server清除日志文件ERRORLOG和删除tempdb.mdf

《SQLServer清除日志文件ERRORLOG和删除tempdb.mdf》数据库再使用一段时间后,日志文件会增大,特别是在磁盘容量不足的情况下,更是需要缩减,以下为缩减方法:如果可以停止SQLSe... 目录缩减 ERRORLOG 文件(停止服务后)停止 SQL Server 服务:找到错误日志文件:删除

mysql删除无用用户的方法实现

《mysql删除无用用户的方法实现》本文主要介绍了mysql删除无用用户的方法实现,文中通过示例代码介绍的非常详细,对大家的学习或者工作具有一定的参考学习价值,需要的朋友们下面随着小编来一起学习学习吧... 1、删除不用的账户(1) 查看当前已存在账户mysql> select user,host,pa

MySQL InnoDB引擎ibdata文件损坏/删除后使用frm和ibd文件恢复数据

《MySQLInnoDB引擎ibdata文件损坏/删除后使用frm和ibd文件恢复数据》mysql的ibdata文件被误删、被恶意修改,没有从库和备份数据的情况下的数据恢复,不能保证数据库所有表数据... 参考:mysql Innodb表空间卸载、迁移、装载的使用方法注意!此方法只适用于innodb_fi

shell脚本自动删除30天以前的文件(最新推荐)

《shell脚本自动删除30天以前的文件(最新推荐)》该文章介绍了如何使用Shell脚本自动删除指定目录下30天以前的文件,并通过crontab设置定时任务,此外,还提供了如何使用Shell脚本删除E... 目录shell脚本自动删除30天以前的文件linux按照日期定时删除elasticsearch索引s

docker如何删除悬空镜像

《docker如何删除悬空镜像》文章介绍了如何使用Docker命令删除悬空镜像,以提高服务器空间利用率,通过使用dockerimage命令结合filter和awk工具,可以过滤出没有Tag的镜像,并将... 目录docChina编程ker删除悬空镜像前言悬空镜像docker官方提供的方式自定义方式总结docker

使用Python在Excel中插入、修改、提取和删除超链接

《使用Python在Excel中插入、修改、提取和删除超链接》超链接是Excel中的常用功能,通过点击超链接可以快速跳转到外部网站、本地文件或工作表中的特定单元格,有效提升数据访问的效率和用户体验,这... 目录引言使用工具python在Excel中插入超链接Python修改Excel中的超链接Python