hdu 3487 Play with Chain

题目链接：http://acm.hdu.edu.cn/showproblem.php?pid=3487

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? 

题意：题意比较简单，就给一个数列两种操作：1，CUT a,b,c，在原数列截取区间[a,b]（a和b指数列a位置和b位置），然后剩下的数列合并在一起，把截取的区间放在那个数列的c　　位置；2，FLIP a b 翻转区间[a,b]。

解法：伸展树+延迟标记。

///*hdu 3487*/
///G++ 687ms  C++ 609ms
#include<iostream>
#include<cstdio>
#include<cstring>
#include<cstdlib>
#include<cmath>
#include<algorithm>
#define inf 0x7fffffff
using namespace std;
const int maxn=3*1e6+10;
struct Splay_Tree
{
    struct Node
    {
        Node *ch[2],*pre;       ///节点的左儿子和右儿子，父亲节点
        int value,size;         ///节点的值和以此节点为根的子树的节点个数
        bool rev;               ///判断是否需要翻转
    }*root,*null,*lb,*rb,node[maxn];
    int count;
    bool flag;
    Node *NewNode(int value,Node *father)///插入新节点
    {
        ///Node *e=node[++count];这样写不行
        Node *e=node+(++count);
        e->value=value;
        e->size=1;
        e->pre=father;
        e->rev=false;
        e->ch[0]=e->ch[1]=null;
        return e;
    }
    void Update(Node *p)///更新节点p的信息
    {
        if (p==null) return;
        p->size=p->ch[0]->size + p->ch[1]->size + 1;
    }
    void PushDown(Node *x)///向下更新标记
    {
        if (x==null) return;
        if (x->rev)
        {
            x->rev=false;
            Node *t=x->ch[0] ;x->ch[0]=x->ch[1] ;x->ch[1]=t;
            x->ch[0]->rev= !x->ch[0]->rev;
            x->ch[1]->rev= !x->ch[1]->rev;
        }
    }
    void Rotate(Node *x,bool d)///右旋转:true;左旋转:false
    {
        Node *y=x->pre;
        y->ch[!d]=x->ch[d];
        if (x->ch[d]!=null) x->ch[d]->pre=y;
        x->pre=y->pre;
        if (y->pre!=null)
        {
            if (y==y->pre->ch[d])
                y->pre->ch[d]=x;
            else y->pre->ch[!d]=x;
        }
        x->ch[d]=y;
        y->pre=x;
        Update(y);
        Update(x);
        if (root==y) root=x;
    }
    void Splay(Node *x,Node *target)///伸展操作
    {
        PushDown(x);
        Node *y;
        while (x->pre!=target)
        {
            y=x->pre;
            if (x==y->ch[0])
            {
                if (y->pre!=target && y->pre->ch[0]==y)
                    Rotate(y,true);
                Rotate(x,true);
            }
            else
            {
                if (y->pre!=target && y->pre->ch[1]==y)
                    Rotate(y,false);
                Rotate(x,false);
            }
        }
        Update(x);
    }
    void Select(int k,Node *y)///选择第k个位置的节点
    {
        Node *x=root;
        PushDown(x);
        while (k!=x->ch[0]->size+1)
        {
            if (k<=x->ch[0]->size)
                x=x->ch[0];
            else
            {
                k -= x->ch[0]->size+1;
                x=x->ch[1];
            }
            PushDown(x);
        }
        Splay(x,y);
    }
    void MakeTree(int l,int r,int *an,Node *p,int side)
    {
        if (l>r) return;
        int mid=(l+r)>>1;
        Node *x;
        x=NewNode(an[mid],p);
        p->ch[side]=x;
        MakeTree(l,mid-1,an,x,0);
        MakeTree(mid+1,r,an,x,1);
        Update(x);
    }
    ///****///
    ///在pos后面插入长度为cnt的区间
    void Insert(int pos,int cnt,int *an)
    {
        Select(pos+1,null);
        Select(pos+2,root);
        MakeTree(1,cnt,an,root->ch[1],0);
        Splay(root->ch[1]->ch[0],null);
    }
    ///删除pos后长度为cnt的区间
    void Delete(int pos,int cnt,int ind)
    {
        Node *p1;
        Select(pos,null);
        Select(pos+cnt+1,root);
        p1=root->ch[1]->ch[0];
        root->ch[1]->ch[0]=null;

        Splay(root->ch[1],null);
        Select(ind+1,null);
        Select(ind+2,root);
        root->ch[1]->ch[0]=p1;
        Splay(root->ch[1],null);///注意不是Splay(root->ch[1]->ch[0],null)
    }
    ///旋转pos后长度为cnt的区间
    void Reverse(int pos,int cnt)
    {
        Select(pos,null);
        Select(pos+cnt+1,root);
        root->ch[1]->ch[0]->rev= !root->ch[1]->ch[0]->rev;
        Splay(root->ch[1]->ch[0],null);
    }
    void Index(int pos)
    {
        Select(pos+1,null);
        printf("%d",root->value);
    }
    void Out()///调试程序
    {
        flag=false;
        Print(root);///打印整个区间
        printf("
");
    }
    void Print(Node *now)///打印区间
    {
        if (now==null) return;
        if (now->ch[0]==null&&now->ch[1]==null)
        {
            if (now->value!=inf)
            {
                if (flag) printf(" ");
                flag=true;
                printf("%d",now->value);
            }
            return;
        }
        PushDown(now);
        Print(now->ch[0]);
        if (now->value!=inf)
        {
            if (flag) printf(" ");
            flag=true;
            printf("%d",now->value);
        }
        Print(now->ch[1]);
    }
    void init()
    {
        count=-1;
        null=0;
        null=NewNode(inf,0);
        null->size=0;
        lb=root=NewNode(inf,null);
        rb=root->ch[1]=NewNode(inf,root);
        Update(root);
    }
}splay;
int an[maxn];
int main()
{
    int n,m;

    char str[5];
    int a,b,c;
    while (scanf("%d%d",&n,&m)!=EOF)
    {
        if (n<0 && m<0) break;
        splay.init();
        for (int i=1 ;i<=n ;i++) an[i]=i;
        splay.Insert(0,n,an);
        while (m--)
        {
            scanf("%s",str);
            if (str[0]=='C')
            {
                scanf("%d%d%d",&a,&b,&c);
                splay.Delete(a,b-a+1,c);
            }
            else
            {
                scanf("%d%d",&a,&b);
                splay.Reverse(a,b-a+1);
            }
        }
        splay.Out();
    }
    return 0;
}