hdu3487 Play with Chain

Play with Chain

Time Limit: 6000/2000 MS (Java/Others)    Memory Limit: 65536/32768 K (Java/Others)

Total Submission(s): 4980    Accepted Submission(s): 2040


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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很显然是Splay维护区间。然而这道题没有自动过滤行末空格，感觉真是坑......（论OIer做ACM题的心态调整2333）

#include<iostream>
#include<cstdio>
#include<cstdlib>
#include<cmath>
#include<cstring>
#include<algorithm>
#define F(i,j,n) for(int i=j;i<=n;i++)
#define D(i,j,n) for(int i=j;i>=n;i--)
#define LL long long
#define pa pair<int,int>
#define MAXN 300005
#define key c[c[rt][1]][0]
using namespace std;
int flag,n,m,rt,x,y,z,c[MAXN][2],s[MAXN],fa[MAXN];
bool rev[MAXN];
char ch[10];
inline int read()
{
int ret=0,flag=1;char ch=getchar();
while (ch<'0'||ch>'9'){if (ch=='-') flag=-1;ch=getchar();}
while (ch>='0'&&ch<='9'){ret=ret*10+ch-'0';ch=getchar();}
return ret*flag;
}
inline void build(int x,int y,int f)
{
if (x>y) return;
int mid=(x+y)>>1;
s[mid]=1;fa[mid]=f;s[mid]=y-x+1;
c[f][mid>f]=mid;
if (x==y) return;
build(x,mid-1,mid);build(mid+1,y,mid);
}
inline void pushup(int k)
{
s[k]=s[c[k][0]]+s[c[k][1]]+1;
}
inline void update(int k)
{
if (!k) return;
rev[k]^=1;
swap(c[k][0],c[k][1]);
}
inline void pushdown(int k)
{
if (!k||!rev[k]) return;
update(c[k][0]);update(c[k][1]);
rev[k]=0;
}
inline int find(int k,int x)
{
pushdown(k);
if (s[c[k][0]]+1==x) return k;
else if (s[c[k][0]]>=x) return find(c[k][0],x);
else return find(c[k][1],x-s[c[k][0]]-1);
}
inline void rotate(int x,int &k)
{
int y=fa[x],z=fa[y],l=(c[y][1]==x),r=l^1;
if (y==k) k=x;
else if (c[z][0]==y) c[z][0]=x;else c[z][1]=x;
fa[x]=z;fa[y]=x;fa[c[x][r]]=y;
c[y][l]=c[x][r];c[x][r]=y;
pushup(y);
}
inline void relax(int x,int k)
{
if (x!=k) relax(fa[x],k);
pushdown(x);
}
inline void splay(int x,int &k)
{
relax(x,k);
while (x!=k)
{
int y=fa[x],z=fa[y];
if (y!=k)
{
if ((c[y][0]==x)^(c[z][0]==y)) rotate(x,k);
else rotate(y,k);
}
rotate(x,k);
}
pushup(x);
}
inline void split(int l,int r)
{
int x=find(rt,l-1),y=find(rt,r+1);
splay(x,rt);splay(y,c[rt][1]);
}
inline void solvecut(int l,int r,int x)
{
split(l,r);
int tmp=key;key=0;pushup(c[rt][1]);pushup(rt);
split(x+1,x);
key=tmp;fa[tmp]=c[rt][1];pushup(c[rt][1]);pushup(rt);
}
inline void solverev(int l,int r)
{
split(l,r);update(key);
pushup(c[rt][1]);pushup(rt);
}
inline void getans(int k)
{
if (!k) return;
pushdown(k);
getans(c[k][0]);
if (k!=1&&k!=n+2)
{
if (flag++) printf(" %d",k-1);
else printf("%d",k-1);
}
getans(c[k][1]);
}
int main()
{
n=read();m=read();
while (n>=0&&m>=0)
{
rt=0;
memset(c,0,sizeof(c));
memset(s,0,sizeof(s));
memset(fa,0,sizeof(fa));
memset(rev,0,sizeof(rev));
build(1,n+2,0);
rt=(n+2)>>1;
F(i,1,m)
{
scanf("%s",ch);x=read()+1;y=read()+1;
if (ch[0]=='C'){z=read()+1;solvecut(x,y,z);}
else solverev(x,y);
}
flag=0;getans(rt);printf("\n");
n=read();m=read();
}
}