hdu3487Play with Chain【splay】

MissZhou要努力發表於2016-04-24
原文網址 : https://blog.csdn.net/zhou_yujia/article/details/51232513
AI

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 




兩種操作：


1.減下來一段按順序貼到某點前面 ：就是木板上delete函式和insert函式和在一起了


2.反轉區間


#include <iostream>
#include <string.h>
#include <stdio.h>
#include <algorithm>
using namespace std;
#define Key_value ch[ch[root][1]][0]
const int MAXN=300009;
const int INF=0x3f3f3f3f;
int pre[MAXN],ch[MAXN][2],key[MAXN],size[MAXN];
int sum[MAXN],rev[MAXN],same[MAXN];
int lx[MAXN],rx[MAXN],mx[MAXN];
int root,tot1;
int s[MAXN],tot2;
int a[MAXN];
int n,q;
//debug部分
void Treavel(int x)
{
    if(x)
    {
        Treavel(ch[x][0]);
        printf("結點%2d:左兒子 %2d 右兒子 %2d 父結點 %2d key=%2d, size= %2d, sum=%2d,rev=%2d same=%2d lx=%2d rx=%2d mx=%2d\n",x,ch[x][0],ch[x][1],pre[x],key[x],size[x],sum[x],rev[x],same[x],lx[x],rx[x],mx[x]);
        Treavel(ch[x][1]);
    }
}
void debug()
{
    printf("root%d\n",root);
    Treavel(root);
}
void NewNode(int &r,int father,int k)
{
    if(tot2)r=s[tot2--];
    else r=++tot1;
    pre[r]=father;
    ch[r][0]=ch[r][1]=0;
    key[r]=k;
    sum[r]=k;
    rev[r]=same[r]=0;
    lx[r]=rx[r]=mx[r]=k;
    size[r]=1;
}
void Update_Same(int r,int v)
{
    if(!r)return;
    key[r]=v;
    sum[r]=v*size[r];
    lx[r]=rx[r]=mx[r]=max(v,v*size[r]);
    same[r]=1;
}
void Update_Rev(int r)
{
    if(!r)return;
    swap(ch[r][0],ch[r][1]);
    swap(lx[r],rx[r]);
    rev[r]^=1;//這裡要注意，一定是異或1
}
void Push_Up(int r)
{
    int lson=ch[r][0],rson=ch[r][1];
    size[r]=size[lson]+size[rson]+1;
    sum[r]=sum[lson]+sum[rson]+key[r];
    lx[r]=max(lx[lson],sum[lson]+key[r]+max(0,lx[rson]));
    rx[r]=max(rx[rson],sum[rson]+key[r]+max(0,rx[lson]));
    mx[r]=max(0,rx[lson])+key[r]+max(0,lx[rson]);
    mx[r]=max(mx[r],max(mx[lson],mx[rson]));
}
void Push_Down(int r)
{
    if(same[r])
    {
        Update_Same(ch[r][0],key[r]);
        Update_Same(ch[r][1],key[r]);
        same[r]=0;
    }
    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,a[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]=pre[root]=size[root]=same[root]=rev[root]=sum[root]=key[root]=0;
    lx[root]=rx[root]=mx[root]=-INF;
    NewNode(root,0,-1);
    NewNode(ch[root][1],root,-1);
    for(int i=1;i<=n;i++)a[i]=i;
    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);
}


//從第pos個數開始連續刪除tot個數
void Delete(int pos,int tot,int pos1)
{
    Splay(Get_Kth(root,pos),0);
    Splay(Get_Kth(root,pos+tot+1),root);
    int tmp=Key_value;
    Key_value=0;
    Push_Up(ch[root][1]);
    Push_Up(root);
    Splay(Get_Kth(root,pos1+1),0);
    Splay(Get_Kth(root,pos1+2),root);
    Key_value=tmp;
    pre[Key_value]=ch[root][1];
    Push_Up(ch[root][1]);
    Push_Up(root);
}

//反轉
void Reverse(int pos,int tot)
{
    Splay(Get_Kth(root,pos),0);
    Splay(Get_Kth(root,pos+tot+1),root);
    Update_Rev(Key_value);
    Push_Up(ch[root][1]);
    Push_Up(root);
}
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 puts("");
    }
    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==-1&&q==-1) 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);
                Delete(x,y-x+1,z);
            }
            else if(op[0]=='F')
            {
                scanf("%d%d",&x,&y);
                Reverse(x,y-x+1);
            }
        }
        cnt=0;
        Inorder(root);
    }
    return 0;
}


            




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