HDU 3487 (Play with Chain-Splay) [template: Splay], hduchain-splay

HDU 3487 (Play with Chain-Splay) [template: Splay], hduchain-splay

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 DescriptionYaoYao 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 wocould 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. cocould you help him?
 
InputThere 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 shoshould not be processed as a case.
 
OutputFor each test case, you shoshould print a line with n numbers. The ith number is the number of the ith diamond on the chain.
Sample Input

8 2CUT 3 5 4FLIP 2 6-1 -1

 
Sample Output

1 4 3 7 6 2 5 8

 
Source2010 ACM-ICPC Multi-University Training Contest (5) -- Host by BJTU
Recommendzhengfeng | We have carefully selected several similar problems for you: 3486 3479 3480 3481

Splay bare question

#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; //rootvoid 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;}

Copyright Disclaimer: This article is an original article by the blogger and cannot be reproduced without the permission of the blogger.