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

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题目大意

就是说给你一串项链，初始时宝石为1−n的顺序，每次两种操作CUT和FLIP，最后项链的样子。

思路

splay直接操作。

代码

#include <cstdio>
#include <cstring>

const int maxn=300000;

int n;

struct splay_tree
{
  int fa[maxn+10],son[2][maxn+10],size[maxn+10],rev[maxn+10],root;

  inline int t(int x)
  {
    return son[1][fa[x]]==x;
  }

  inline int push_rev(int x)
  {
    rev[x]^=1;
    int t=son[0][x];
    son[0][x]=son[1][x];
    son[1][x]=t;
    return 0;
  }

  inline int pushdown(int x)
  {
    if(rev[x])
      {
        if(son[0][x])
          {
            push_rev(son[0][x]);
          }
        if(son[1][x])
          {
            push_rev(son[1][x]);
          }
        rev[x]=0;
      }
    return 0;
  }

  inline int updata(int x)
  {
    return size[x]=size[son[0][x]]+size[son[1][x]]+1;
  }

  inline int rotate(int x)
  {
    int k=t(x),f=fa[x];
    if(fa[f])
      {
        son[t(f)][fa[f]]=x;
      }
    fa[x]=fa[f];
    if(son[!k][x])
      {
        fa[son[!k][x]]=f;
      }
    son[k][f]=son[!k][x];
    fa[f]=x;
    son[!k][x]=f;
    updata(f);
    updata(x);
    return 0;
  }

  inline int splay(int x,int c)
  {
    while(fa[x]!=c)
      {
        int f=fa[x];
        if(fa[f]==c)
          {
            rotate(x);
          }
        else if(t(x)==t(f))
          {
            rotate(f);
            rotate(x);
          }
        else
          {
            rotate(x);
            rotate(x);
          }
      }
    if(!c)
      {
        root=x;
      }
    return 0;
  }

  inline int getkth(int x)
  {
    int now=root;
    while(now)
      {
        pushdown(now);
        if(size[son[0][now]]+1==x)
          {
            return now;
          }
        else if(size[son[0][now]]+1<x)
          {
            x-=size[son[0][now]]+1;
            now=son[1][now];
          }
        else
          {
            now=son[0][now];
          }
      }
    return 0;
  }

  int build(int l,int r)
  {
    int mid=(l+r)>>1;
    rev[mid]=0;
    size[mid]=r-l+1;
    if(l<=mid-1)
      {
        son[0][mid]=build(l,mid-1);
        fa[son[0][mid]]=mid;
      }
    if(mid+1<=r)
      {
        son[1][mid]=build(mid+1,r);
        fa[son[1][mid]]=mid;
      }
    return mid;
  }

  inline int cut(int l,int r,int x)
  {
    int now;
    if(l==1)
      {
        if(r==n)
          {
            now=root;
            root=0;
          }
        else
          {
            int w=getkth(r+1);
            splay(w,0);
            now=son[0][w];
            son[0][w]=0;
            fa[now]=0;
            updata(w);
          }
      }
    else
      {
        int w=getkth(l-1);
        splay(w,0);
        if(r==n)
          {
            now=son[1][w];
            son[1][w]=0;
            fa[now]=0;
          }
        else
          {
            int y=getkth(r+1);
            splay(y,w);
            now=son[0][y];
            son[0][y]=0;
            fa[now]=0;
            updata(y);
          }
        updata(w);
      }
    if(!x)
      {
        if(!root)
          {
            root=now;
          }
        else
          {
            int w=getkth(1);
            splay(w,0);
            son[0][w]=now;
            fa[now]=w;
            updata(w);
          }
      }
    else
      {
        int w=getkth(x);
        splay(w,0);
        if(x==n-(r-l+1))
          {
            son[1][w]=now;
            fa[now]=w;
          }
        else
          {
            int y=getkth(x+1);
            splay(y,w);
            son[0][y]=now;
            fa[now]=y;
            updata(y);
          }
        updata(w);
      }
    return 0;
  }

  inline int reverse(int l,int r)
  {
    if(l==1)
      {
        if(r==n)
          {
            push_rev(root);
          }
        else
          {
            int x=getkth(r+1);
            splay(x,0);
            push_rev(son[0][x]);
          }
      }
    else
      {
        int x=getkth(l-1);
        splay(x,0);
        if(r==n)
          {
            push_rev(son[1][x]);
          }
        else
          {
            int y=getkth(r+1);
            splay(y,x);
            push_rev(son[0][y]);
          }
      }
    return 0;
  }

  inline int mem()
  {
    memset(fa,0,sizeof fa);
    memset(son,0,sizeof son);
    memset(size,0,sizeof size);
    memset(rev,0,sizeof rev);
    return 0;
  }
};

splay_tree st;
int m,a,b,c;
char s[10];

int main()
{
  while(1)
    {
      st.mem();
      scanf("%d%d",&n,&m);
      if((n<0)&&(m<0))
        {
          break;
        }
      st.build(1,n);
      st.root=(n+1)>>1;
      while(m--)
        {
          scanf("%s%d%d",s,&a,&b);
          if(s[0]=='C')
            {
              scanf("%d",&c);
              st.cut(a,b,c);
            }
          else
            {
              st.reverse(a,b);
            }
        }
      for(register int i=1; i<n; ++i)
        {
          printf("%d ",st.getkth(i));
        }
      printf("%d\n",st.getkth(n));
    }
  return 0;
}