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UVA 11922 Splay区间翻转+分裂+合并

 - Permutation Transformer
Time Limit:2000MS     Memory Limit:0KB     64bit IO Format:%lld & %llu
Submit Status Practice UVA 11922
Appoint description: 

Description

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  Permutation Transformer 

Write a program to transform the permutation 1, 2, 3,..., n according to m instructions. Each instruction (ab) means to take out the subsequence from the a-th to the b-th element, reverse it, then append it to the end.

Input 

There is only one case for this problem. The first line contains two integers n and m ( 1$ \le$nm$ \le$100, 000). Each of the next m lines contains an instruction consisting of two integers a and b ( 1$ \le$a$ \le$b$ \le$n).

Output 

Print n lines, one for each integer, the final permutation.


Explanation of the sample below

Instruction (2,5): Take out the subsequence {2,3,4,5}, reverse it to {5,4,3,2}, append it to the remaining permutation {1,6,7,8,9,10}

Instruction (4,8): The subsequence from the 4-th to the 8-th element of {1,6,7,8,9,10,5,4,3,2} is {8,9,10,5,4}. Take it out, reverse it, and you‘ll get the sample output.


Warning: Don‘t use cincout for this problem, use faster i/o methods e.g scanfprintf.

Sample Input 

10 2
2 5
4 8

Sample Output 

1
6
7
3
2
4
5
10
9
8





对序列进行m次操作,每次将[L,R]区间翻转后移到尾部。

提取区间之后翻转,分裂,进行两次合并


代码:

/*************************************************************************
    > File Name: Spaly.cpp
    > Author: acvcla
    > QQ:
    > Mail: acvcla@gmail.com
    > Created Time: 2014Äê11ÔÂ16ÈÕ ÐÇÆÚÈÕ 00ʱ14·Ö26Ãë
 ************************************************************************/
#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;
const int maxn = 2e5 + 100;
const int inf=1e9+7;
#define rep(i,a,b) for(int i=(a);i<=(b);i++)
#define pb push_back
int ch[maxn][2],pre[maxn],rev[maxn],siz[maxn];
int root,tot;
int key[maxn];
void newnode(int &x,int fa,int k)
{
	x=++tot;
	key[x]=k;
	siz[x]=1;
	pre[x]=fa;
	rev[x]=ch[x][1]=ch[x][0]=0;
}
void push_up(int x){
	if(!x)return;
	siz[x]=siz[ch[x][0]]+siz[ch[x][1]]+1;
}
void upadte_rev(int x)
{
	if(!x)return ;
	swap(ch[x][0],ch[x][1]);
	rev[x]^=1;
}
void push_down(int x)
{
	if(!x)return;
	if(rev[x]){
		upadte_rev(ch[x][0]);
		upadte_rev(ch[x][1]);
		rev[x]^=1;
	}
}
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;
}
void built(int &x,int L,int R,int fa)
{
	if(L>R)return;
	int mid=(R+L)>>1;
	newnode(x,fa,mid);
	built(ch[x][0],L,mid-1,x);
	built(ch[x][1],mid+1,R,x);
	push_up(x);
}
void init(int n)
{
	siz[0]=pre[0]=0;
	root=tot=ch[0][0]=ch[0][1]=0;
	newnode(root,0,inf);
	newnode(ch[root][1],root,inf);
	built(ch[ch[root][1]][0],1,n,ch[root][1]);
	push_up(ch[root][1]);
	push_up(root);
}
int Get_kth(int x,int k){
	push_down(x);
	int sz=siz[ch[x][0]]+1;
	if(sz==k)return x;
	if(sz>k)return Get_kth(ch[x][0],k);
	return Get_kth(ch[x][1],k-sz);
}
void reverse(int L,int R)
{
	Splay(Get_kth(root,L-1),0);
	Splay(Get_kth(root,R+1),root);
	upadte_rev(ch[ch[root][1]][0]);
}
void append_to_end()
{
	int t=ch[ch[root][1]][0];
	ch[ch[root][1]][0]=0;
	push_up(ch[root][1]);
	push_up(root);
	Splay(Get_kth(root,siz[root]-1),0);
	Splay(Get_kth(root,siz[root]),root);
	ch[ch[root][1]][0]=t;
	pre[t]=ch[root][1];
	push_up(ch[root][1]);
	push_up(root);
}
void print(int x){
	if(!x)return;
	push_down(x);
	print(ch[x][0]);
	if(key[x]!=inf)printf("%d\n",key[x]);
	print(ch[x][1]);
}
int main(int argc, char const *argv[])
{
		int n,m,L,R;
		while(~scanf("%d%d",&n,&m))
		{
			init(n);
			while(m--){
				scanf("%d%d",&L,&R);
				reverse(L+1,R+1);
				append_to_end();
			}
			print(root);
		}
		return 0;
}



UVA 11922 Splay区间翻转+分裂+合并