HDU--3487 Play with Chain (Splay伸展树)

Play with Chain

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

题意：FLIP操作就是区间的翻转，而CUT则是把某个区间移到某个点后面，本来打算用几次翻转来合成CUT函数，最后没有推出来。so借鉴kuangbin巨巨的作法

#include <cstdio>
#include <cstdlib>
#include <algorithm>
using namespace std;
const int maxn = 3e5+10;
int siz[maxn],pre[maxn],ch[maxn][2],rev[maxn],key[maxn];
int tot,root,n,m;
void NewNode(int &r,int father,int k)
{
    r = ++tot;
    pre[r] = father;
    ch[r][0] = ch[r][1] = 0;
    key[r] = k;
    siz[r] = 1;
    rev[r] = 0;
}
void update_rev(int r)
{
    if (!r)
        return ;
    swap(ch[r][0],ch[r][1]);
    rev[r] ^= 1;
}
void push_up(int r)
{
    siz[r] = siz[ch[r][0]] + siz[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) >> 1;
    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 = tot = 0;
    NewNode(root,0,-1);
    NewNode(ch[root][1],root,-1);
    build(ch[ch[root][1]][0],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
        {
            int y = pre[r];
            int kind = (ch[pre[y]][1] == y);
            push_down(pre[y]);
            push_down(y);
            push_down(r);
            if (ch[y][kind] == r)
            {
                Rotate(y,!kind);
                Rotate(r,!kind);
            }
            else
            {
                Rotate(r,kind);
                Rotate(r,!kind);
            }
        }
    }
    push_up(r);
    if (goal == 0)
        root = r;
}
int Get_kth(int r,int k)
{
    push_down(r);
    int t = siz[ch[r][0]] + 1;
    if (k == t)
        return r;
    if (k >= t)
        return Get_kth(ch[r][1],k-t);
    else
        return Get_kth(ch[r][0],k);
}
void Reverse(int u,int v)
{
    if (u > v)
        return;
    Splay(Get_kth(root,u),0);
    Splay(Get_kth(root,v+2),root);
    update_rev(ch[ch[root][1]][0]);
    push_up(ch[root][1]);
    push_up(root);
}
void cut(int x,int y,int z)          // cut函数本来想着用几次翻转来实现，推了半天没有推出来。so借鉴kuangbin巨巨的做法
{
    Splay(Get_kth(root,x),0);
    Splay(Get_kth(root,y+2),root);
    int tmp = ch[ch[root][1]][0];
    ch[ch[root][1]][0] = 0;
    push_up(ch[root][1]);
    push_up(root);
    Splay(Get_kth(root,z+1),0);
    Splay(Get_kth(root,z+2),root);
    ch[ch[root][1]][0] = tmp;
    pre[ch[ch[root][1]][0]] = ch[root][1];
    push_up(ch[root][1]);
    push_up(root);
}
bool flag;
void dfs(int r)
{
    if (!r)
        return;
    push_down(r);
    dfs(ch[r][0]);
    if (r != -1 && key[r] != -1)
    {
        printf(flag ? " %d":"%d",key[r]);
        flag = 1;
    }
    dfs(ch[r][1]);
}
int main(void)
{
    #ifndef ONLINE_JUDGE
        freopen("in.txt","r",stdin);
    #endif
    while (~scanf ("%d%d",&n,&m))
    {
        if (n < 0 && m < 0)
            break;
        init();
        for (int i = 0; i < m; i++)
        {
            char op[6];
            int x,y,z;
            scanf ("%s",op);
            if (op[0] == 'C')
            {
                scanf ("%d%d%d",&x,&y,&z);
                cut(x,y,z);
            }
            else
            {
                scanf ("%d%d",&x,&y);
                Reverse(x,y);
            }
        }
        flag = 0;
        dfs(root);
        printf("\n");
    }
    return 0;
}