题目:
http://acm.hdu.edu.cn/showproblem.php?pid=1890
题意:
某台机器要排序一个序列,然而这台机器只能实现一种操作,将序列中某段区间完全翻转,某君提出了个思路,对长度为n的序列,机器只要翻转n次,第i次找到[i,n]中最小的数x的位置k,然后翻转[i,k]即可(想不明白为什么这么翻就可以的请看样例去),求输出每次翻转前的序列中,x的位置
思路:
在学伸展树,啊不得不说学会一种新数据结构的心情真是爽爆了,做这题前我拿poj3468先练了练手(那是道线段树加懒标记的经典问题,然而也可以用伸展树搞),伸展树在处理这种区间翻转的序列问题真是得心应手
用数组下标建splay树,当我们需要翻转[i,k]区间时,先将i-1结点旋转为根节点,然后将k+1结点旋转为i-1结点的右儿子,这时[i,k]区间就是i-1结点的左子树(这种处理方法非常非常经典,想不明白自己画个图看看,论文里面也有),此时对i-1结点的左子树打上翻转标记即可
本题的详细算法为,每次找到x,将x旋转至根节点的右儿子的左儿子(我没打错!看看上一段),此时x的左子树的结点数即为序列翻转前x前的数字数量,然后给x的左子树打上翻转标记,删去x结点,合并x的左右子树,循环执行这一步骤即可
为了方便处理,在序列的前后各加一个结点
代码:
#include <iostream>
#include <stdio.h>
#include <algorithm>
using namespace std;
const int MAXSIZE = 1e5+100;
struct Node{
int key, sz;
Node *ch[2], *pnt;
bool rever;
inline void push_up(){
sz = 1;
if (ch[0]) sz += ch[0]->sz;
if (ch[1]) sz += ch[1]->sz;
}
inline void push_down(){
if (!rever) return;
swap(ch[0],ch[1]);
if (ch[0]) ch[0]->rever = !ch[0]->rever;
if (ch[1]) ch[1]->rever = !ch[1]->rever;
rever = false;
}
};
Node *root;
struct node{
int id, num;
};
bool operator <(const node&p1, const node&p2){
if (p1.num!=p2.num) return p1.num<p2.num;
else return p1.id<p2.id;
}
node arr[MAXSIZE];
void init(){
root = NULL;
}
void Treaval(Node *x) {
if(x) {
Treaval(x->ch[0]);
cout<<"key: "<<x->key
<<" sz: "<<x->sz
<<" reverse: "<<x->rever;
if (x->ch[0]) cout<<" lchild: "<<x->ch[0]->key;
else cout<<" lchild: null";
if (x->ch[1]) cout<<" rchild: "<<x->ch[1]->key;
else cout<<" rchild: null";
if (x->pnt) cout<<" pnt: "<<x->pnt->key;
else cout<<" pnt: null";
cout<<endl;
Treaval(x->ch[1]);
}
}
void debug() {printf("root: %d\n",root->key);Treaval(root);}
void srotate(Node *x, bool d){
Node *y = x->pnt;
y->ch[!d] = x->ch[d];
if (x->ch[d] != NULL)
x->ch[d]->pnt = y;
x->pnt = y->pnt;
if (y->pnt != NULL){
if (y == y->pnt->ch[d])
y->pnt->ch[d] = x;
else
y->pnt->ch[!d] = x;
}
x->ch[d] = y;
y->pnt = x;
y->push_up();
x->push_up();
}
void splay(Node *x, Node *target){
Node *y;
while (x->pnt != target){
y = x->pnt;
if (y->pnt) y->pnt->push_down();
y->push_down();
x->push_down();
if (x == y->ch[0]){
if (y->pnt != target && y == y->pnt->ch[0])
srotate(y, true);
srotate(x, true);
}
else{
if (y->pnt != target && y == y->pnt->ch[1])
srotate(y, false);
srotate(x, false);
}
}
if (target == NULL)
root = x;
}
Node* sinsert(int key){//插入一个值
if (root == NULL){
root = new Node;
root->ch[0] = root->ch[1] = root->pnt = NULL;
root->rever = false;
root->key = key;
root->sz = 1;
return root;
}
Node *x = root, *y;
while (1){
x->sz++;
if (key < x->key){
if (x->ch[0] != NULL)
x = x->ch[0];
else{
x->ch[0] = new Node;
y = x->ch[0];
y->key = key;
y->sz = 1;
y->ch[0] = y->ch[1] = NULL;
y->rever = false;
y->pnt = x;
break;
}
}
else{
if (x->ch[1] != NULL)
x = x->ch[1];
else{
x->ch[1] = new Node;
y = x->ch[1];
y->key = key;
y->sz = 1;
y->ch[0] = y->ch[1] = NULL;
y->rever = false;
y->pnt = x;
break;
}
}
}
splay(y, NULL);
return y;
}
Node* searchMinimum(Node *x){
Node *y = x->pnt;
if (x) x->push_down();
while (x->ch[0]){
x = x->ch[0];
if (x) x->push_down();
}
splay(x,y);
return x;
}
Node* searchMaximum(Node *x){
Node *y = x->pnt;
while (x->ch[1])
x = x->ch[1];
splay(x,y);
return x;
}
//执行join时,需要保证x子树的任一结点均小于y中任一结点
Node* join(Node *s1, Node *s2){
if (!s1) return s2;
s1->rever = !s1->rever;
if (!s2) return s1;
Node *t = searchMinimum(s2);
t->ch[0] = s1;
s1->pnt = t;
t->push_up();
return t;
}
void del(Node *x){
if (x->ch[0]) x->ch[0]->pnt = NULL;
if (x->ch[1]) x->ch[1]->pnt = NULL;
root = join(x->ch[0],x->ch[1]);
delete x;
}
Node* arrmap[MAXSIZE];
int main(){
int n;
while (scanf("%d",&n),n!=0){
init();
for (int i=0;i<n;++i){
scanf("%d",&arr[i].num);
arr[i].id = i+1;
}
sort(arr,arr+n);
for (int i=0;i<n;++i){
//cout<<arr[i].id<<" "<<arr[i].num<<endl;
arrmap[i] = sinsert(arr[i].id);
}
//debug();
for (int i=0;i<n-1;++i){
int ans = i + 1;
arrmap[i]->push_down();
splay(arrmap[i],NULL);
if (root->ch[0]) ans += root->ch[0]->sz;
//debug();
printf("%d ",ans);
del(root);
//debug();
}
printf("%d\n",n);
}
return 0;
}