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HDU3487(splay区间翻转+区间切割）

2024-08-04 00:22:46   225人阅读



F - Play with Chain
Time Limit:2000MS     Memory Limit:32768KB     64bit IO Format:%I64d & %I64u
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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
 


题意：开始有一个1，2,3，。。。n的序列，进行m次操作，cut a b c将区间[a,b]取出得到新序列，将区间插入到新序列第c个元素之后。filp a b 将区间a，b翻转，输出最终的序列。

思路：对于cut操作我们需要先提取出区间[a,b]然后，先暂时分裂出去，然后以c为边界分裂左右两部分，然后合并左边的和区间[a,b]，将最大的旋转至根，然后和右边的合并。对于filp操作，我们可以使用lazy标记，先不翻转，需要的时候再翻转，代码如下

/*************************************************************************
    > File Name: A.cpp
    > Author: acvcla
    > QQ:
    > Mail: acvcla@gmail.com 
    > Created Time: 2014年11月17日 星期一 23时34分13秒
 ************************************************************************/
#include<iostream>
#include<algorithm>
#include<cstdio>
#include<vector>
#include<cstring>
#include<map>
#include<queue>
#include<stack>
#include<string>
#include<cstdlib>
#include<ctime>
#include<set>
#include<math.h>
using namespace std;
typedef long long LL;
typedef pair<int,int>pii;
const int maxn = 3e5 + 10;
#define rep(i,a,b) for(int i=(a);i<=(b);i++)
#define pb push_back
int siz[maxn],rev[maxn],pre[maxn],ch[maxn][2],key[maxn],ans[maxn];
int tot,root,n;
inline void newnode(int &x,int fa,int k){
	x=++tot;
	pre[x]=fa;
	siz[x]=1;
	key[x]=k;
	ch[x][0]=ch[x][1]=rev[x]=0;
}
inline void Modify(int x){
	if(!x)return;
	rev[x]^=1;
}
inline void push_down(int x){
	if(x&&rev[x]){
		swap(ch[x][0],ch[x][1]);
		Modify(ch[x][0]);
		Modify(ch[x][1]);
		Modify(x);
	}
}
inline void push_up(int x){
	if(!x)return ;
	siz[x]=siz[ch[x][0]]+siz[ch[x][1]]+1;
}
void built(int &x,int L,int R,int fa){
	if(L>R)return;
	int M=(L+R)>>1;
	newnode(x,fa,M);
	built(ch[x][0],L,M-1,x);
	built(ch[x][1],M+1,R,x);
	push_up(x);
}
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;
	ch[x][kind]=y;

	if(pre[y])ch[pre[y]][ch[pre[y]][1]==y]=x;
	pre[x]=pre[y];
	pre[y]=x;

	push_up(y);
	push_up(x);
}
void Splay(int x,int goal){
	while(pre[x]!=goal){
		if(pre[pre[x]]==goal){
			Rotate(x,ch[pre[x]][0]==x);
		}else{
			int y=pre[x];
			int kind=(ch[pre[y]][0]==y);
			if(ch[y][kind]==x){
				Rotate(x,!kind);
				Rotate(x,kind);
			}else{
				Rotate(y,kind);
				Rotate(x,kind);
			}
		}
	}
	if(goal==0)root=x;
}
int Get_kth(int x,int k){
	push_down(x);
	int tz=siz[ch[x][0]]+1;
	if(tz==k)return x;
	if(tz>k)return Get_kth(ch[x][0],k);
	return Get_kth(ch[x][1],k-tz);
}
void init(int n){
	root=tot=siz[0]=pre[0]=ch[0][0]=ch[0][1]=rev[0]=0;
	newnode(root,0,-1);
	newnode(ch[root][1],root,n+1);
	built(ch[ch[root][1]][0],1,n,ch[root][1]);
	push_up(ch[root][1]);
	push_up(root);
}
int Get_max(int x){
	push_down(x);
	while(ch[x][1]){
		x=ch[x][1];
		push_down(x);
	}
	return x;
}
void merge(int root1,int root2)/*root2接到root1右子树，要求root1无右子树*/
{
	ch[root1][1]=root2;
	pre[root2]=root1;
}
int __;
void travel(int x){
	if(!x)return;
	push_down(x);
	travel(ch[x][0]);
	ans[__++]=key[x];
	travel(ch[x][1]);
}
int main(){
		int n,m;
		while(~scanf("%d%d",&n,&m)){
			if(n<0&&m<0)return 0;
			init(n);
			char op[10];
			int L,R,C;
			while(m--){
				scanf("%s%d%d",op,&L,&R);
				if(op[0]=='F'){
					Splay(Get_kth(root,L),0);
					Splay(Get_kth(root,R+2),root);
					Modify(ch[ch[root][1]][0]);
				}else{
					scanf("%d",&C);
					Splay(Get_kth(root,L),0);
					Splay(Get_kth(root,R+2),root);
					int root1=ch[ch[root][1]][0];/*删除区间[L,R]*/
					ch[ch[root][1]][0]=0;
					push_up(ch[root][1]);
					push_up(root);
					Splay(Get_kth(root,C+1),0);/*先分裂区间C两边，插入区间[L,R]，然后合并*/
					int root2=ch[root][1];
					merge(root,root1);
					push_up(root);
					Splay(Get_max(root),0);
					merge(root,root2);
					push_up(root);
				}
			}
			__=0;
			travel(root);
			rep(i,1,n)printf("%d%c",ans[i],i==n?'\n':' ');
		}
		return 0;
}


HDU3487(splay区间翻转+区间切割）


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