Play with Chain
Time Limit: 6000/2000 MS (Java/Others) Memory Limit: 65536/32768 K (Java/Others)Total Submission(s): 4679 Accepted Submission(s): 1892
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?
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.
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
Recommend
Splay 裸题
#include<cstdio>
#include<cstring>
#include<cstdlib>
#include<algorithm>
#include<functional>
#include<iostream>
#include<cmath>
#include<cctype>
#include<ctime>
using namespace std;
#define For(i,n) for(int i=1;i<=n;i++)
#define Fork(i,k,n) for(int i=k;i<=n;i++)
#define Rep(i,n) for(int i=0;i<n;i++)
#define ForD(i,n) for(int i=n;i;i--)
#define RepD(i,n) for(int i=n;i>=0;i--)
#define Forp(x) for(int p=pre[x];p;p=next[p])
#define Forpiter(x) for(int &p=iter[x];p;p=next[p])
#define Lson (x<<1)
#define Rson ((x<<1)+1)
#define MEM(a) memset(a,0,sizeof(a));
#define MEMI(a) memset(a,127,sizeof(a));
#define MEMi(a) memset(a,128,sizeof(a));
#define INF (2139062143)
#define F (100000007)
#define MAXN (300000+10)
#define MAXM (300000+10)
typedef long long ll;
ll mul(ll a,ll b){return (a*b)%F;}
ll add(ll a,ll b){return (a+b)%F;}
ll sub(ll a,ll b){return (a-b+(a-b)/F*F+F)%F;}
void upd(ll &a,ll b){a=(a%F+b%F)%F;}
int n,m;
class Splay
{
public:
int father[MAXN],siz[MAXN],n;
int ch[MAXN][2],val[MAXN];
bool root[MAXN],rev[MAXN];
int roo; //root
void mem(int _n)
{
MEM(father) MEM(siz) MEM(root) MEM(rev) MEM(ch) MEM(val) flag=0;
n=0;
roo=1;
build(roo,1,_n,0);root[1]=1;
}
void newnode(int &x,int f,int v)
{
x=++n;
father[x]=f;
val[x]=v-1;
}
void build(int &x,int L,int R,int f)
{
if (L>R) return ;
int m=(L+R)>>1;
newnode(x,f,m);
build(ch[x][0],L,m-1,x);
build(ch[x][1],m+1,R,x);
maintain(x);
}
int getkth(int x,int k)
{
pushdown(x);
int t;
if (ch[x][0]) t=siz[ch[x][0]]; else t=0;
if (t==k-1) return x;
else if (t>=k) return getkth(ch[x][0],k);
else return getkth(ch[x][1],k-t-1);
}
void pushdown(int x)
{
if (x) if (rev[x])
{
swap(ch[x][0],ch[x][1]);
if (ch[x][0]) rev[ ch[x][0] ]^=1;
if (ch[x][1]) rev[ ch[x][1] ]^=1;
rev[x]^=1;
}
}
void maintain(int x)
{
siz[x]=siz[ch[x][0]]+siz[ch[x][1]]+1;
}
void rotate(int x)
{
int y=father[x],kind=ch[y][1]==x;
pushdown(y); pushdown(x);
ch[y][kind]=ch[x][!kind];
if (ch[y][kind]) {
father[ch[y][kind]]=y;
}
father[x]=father[y];
father[y]=x;
ch[x][!kind]=y;
if (root[y])
{
root[x]=1;root[y]=0;roo=x;
}
else
{
ch[father[x]][ ch[father[x]][1]==y ] = x;
}
maintain(y);maintain(x);
}
void splay(int x)
{
while(!root[x])
{
int y=father[x];
int z=father[y];
if (root[y]) rotate(x);
else if ( (ch[y][1]==x)^(ch[z][1]==y) )
{
rotate(x); rotate(x);
}
else
{
rotate(y); rotate(x);
}
}
roo=x;
}
void splay(int x,int r)
{
while(!(father[x]==r))
{
int y=father[x];
int z=father[y];
if (father[y]==r) rotate(x);
else if ( (ch[y][1]==x)^(ch[z][1]==y) )
{
rotate(x); rotate(x);
}
else
{
rotate(y); rotate(x);
}
}
}
void Cut(int a,int b,int c)
{
int x=getkth(roo,a),y=getkth(roo,b);
splay(x);
splay(y,roo);
pushdown(x);pushdown(y);
int z=ch[y][0];
ch[y][0]=0; maintain(y); maintain(x);
int u=getkth(roo,c),v=getkth(roo,c+1);
splay(u);
splay(v,roo);
pushdown(u);pushdown(v);
ch[v][0]=z;father[z]=v;
maintain(v);maintain(u);
}
void Flip(int a,int b)
{
int x=getkth(roo,a),y=getkth(roo,b);
splay(x);
splay(y,roo);
pushdown(x);pushdown(y);
int z=ch[y][0];
rev[z]^=1;
maintain(y); maintain(x);
}
bool flag;
void print(int x)
{
if (x==0) return ;
pushdown(x);
print(ch[x][0]);
if (val[x]!=0&&val[x]!=n-1)
{
if (flag) putchar(' '); else flag=1;
printf("%d",val[x]);
}
print(ch[x][1]);
}
}S;
char s[MAXN];
int main()
{
// freopen("hdu3487.in","r",stdin);
// freopen(".out","w",stdout);
while(cin>>n>>m)
{
if (n<0&&m<0) break;
n+=2;
S.mem(n);
For(i,m)
{
scanf("%s",s);
if (s[0]=='C')
{
int a,b,c;
scanf("%d%d%d",&a,&b,&c);
S.Cut(a,b+2,c+1);
} else {
int a,b;
scanf("%d%d",&a,&b);
S.Flip(a,b+2);
}
}
S.print(S.roo);cout<<endl;
}
return 0;
}