HDU 3487 Play with Chain （splay tree)

Play with Chain

Time Limit: 6000/2000 MS (Java/Others)    Memory Limit: 65536/32768 K (Java/Others)
Total Submission(s): 2783    Accepted Submission(s): 1141

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

 

Source

2010 ACM-ICPC Multi-University Training Contest（5）——Host by BJTU

 

Recommend

zhengfeng

/* ***********************************************
Author        :kuangbin
Created Time  :2013/8/25 23:35:30
File Name     :F:2013ACM练习专题学习splay_tree_2HDU3487.cpp
************************************************ */

#include <stdio.h>
#include <string.h>
#include <iostream>
#include <algorithm>
#include <vector>
#include <queue>
#include <set>
#include <map>
#include <string>
#include <math.h>
#include <stdlib.h>
#include <time.h>
using namespace std;

#define Key_value ch[ch[root][1]][0]
const int MAXN = 300010;
int pre[MAXN],ch[MAXN][2],root,tot1;
int key[MAXN],rev[MAXN];
int size[MAXN];
int s[MAXN],tot2;
int n,q;

void NewNode(int &r,int father,int k)
{
    if(tot2) r = s[tot2--];
    else r = ++tot1;
    key[r] = k;
    pre[r] = father;
    rev[r] = 0;
    ch[r][0] = ch[r][1] = 0;
    size[r] = 1;
}
//反转的更新
void Update_Rev(int r)
{
    if(!r) return;
    swap(ch[r][0],ch[r][1]);
    rev[r] ^= 1;
}
void push_up(int r)
{
    size[r] = size[ch[r][0]] + size[ch[r][1]] + 1;
}
void push_down(int r)
{
    if(rev[r])
    {
        Update_Rev(ch[r][0]);
        Update_Rev(ch[r][1]);
        rev[r] = 0;
    }
}

void Build(int &x,int l,int r,int father)
{
    if(l > r)return;
    int mid = (l+r)/2;
    NewNode(x,father,mid);
    Build(ch[x][0],l,mid-1,x);
    Build(ch[x][1],mid+1,r,x);
    push_up(x);
}
void Init()
{
    root = tot1 = tot2 = 0;
    ch[root][0] = ch[root][1] = size[root] = key[root] = pre[root] = rev[root] = 0;
    NewNode(root,0,-1);
    NewNode(ch[root][1],root,-1);
    Build(Key_value,1,n,ch[root][1]);
    push_up(ch[root][1]);
    push_up(root);
}
void Rotate(int x,int kind)
{
    int y = pre[x];
    push_down(y);
    push_down(x);
    ch[y][!kind] = ch[x][kind];
    pre[ch[x][kind]] = y;
    if(pre[y])
        ch[pre[y]][ch[pre[y]][1]==y] = x;
    pre[x] = pre[y];
    ch[x][kind] = y;
    pre[y] = x;
    push_up(y);
}
void Splay(int r,int goal)
{
    push_down(r);
    while(pre[r] != goal)
    {
        if(pre[pre[r]] == goal)
        {
            push_down(pre[r]);
            push_down(r);
            Rotate(r,ch[pre[r]][0]==r);
        }
        else
        {
            push_down(pre[pre[r]]);
            push_down(pre[r]);
            push_down(r);
            int y = pre[r];
            int kind = ch[pre[y]][0]==y;
            if(ch[y][kind] == r)
            {
                Rotate(r,!kind);
                Rotate(r,kind);
            }
            else
            {
                Rotate(y,kind);
                Rotate(r,kind);
            }
        }
    }
    push_up(r);
    if(goal == 0)root = r;
}
int Get_kth(int r,int k)
{
    push_down(r);
    int t = size[ch[r][0]] + 1;
    if(t == k)return r;
    if(t > k)return Get_kth(ch[r][0],k);
    else return Get_kth(ch[r][1],k-t);
}
//将[l,r]段剪切下来，放在剩下段的第c个后面
void CUT(int l,int r,int c)
{
    Splay(Get_kth(root,l),0);
    Splay(Get_kth(root,r+2),root);
    int tmp = Key_value;
    Key_value = 0;
    push_up(ch[root][1]);
    push_up(root);
    Splay(Get_kth(root,c+1),0);
    Splay(Get_kth(root,c+2),root);
    Key_value = tmp;
    pre[Key_value] = ch[root][1];
    push_up(ch[root][1]);
    push_up(root);
}
//将[l,r]段反转
void FLIP(int l,int r)
{
    Splay(Get_kth(root,l),0);
    Splay(Get_kth(root,r+2),root);
    Update_Rev(Key_value);
    push_up(ch[root][1]);
    push_up(root);
}
//按顺序输出
int cnt;
void InOrder(int r)
{
    if(!r)return;
    push_down(r);
    InOrder(ch[r][0]);
    if(cnt >=1 && cnt <= n)
    {
        printf("%d",key[r]);
        if(cnt < n)printf(" ");
        else printf("
");
    }
    cnt++;
    InOrder(ch[r][1]);
}
int main()
{
    //freopen("in.txt","r",stdin);
    //freopen("out.txt","w",stdout);
    while(scanf("%d%d",&n,&q) == 2)
    {
        if( n < 0 && q < 0)break;
        Init();
        char op[20];
        int x,y,z;
        while(q--)
        {
            scanf("%s",op);
            if(op[0] == 'C')
            {
                scanf("%d%d%d",&x,&y,&z);
                CUT(x,y,z);
            }
            else
            {
                scanf("%d%d",&x,&y);
                FLIP(x,y);
            }
        }
        cnt = 0;
        InOrder(root);
    }
    return 0;
}