#HDU3487#Play the Chain

Play with Chain

Time Limit: 6000/2000 MS (Java/Others)    Memory Limit: 65536/32768 K (Java/Others)
Total Submission(s): 6438    Accepted Submission(s): 2569

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

 

上周学了AVl平衡树，作为一个好学生，坚持不copy，每次都从空文件开始写的我感觉AVl代码真的是太，，，长了。

这周一来老师就讲了splay，给了一种很巧妙的简洁写法，个人感觉非常好，并且不容易写错，没这么多细节注意。（手残的人尤其开心）

splay的操作跟avl很像，只是基础操作单位和操作顺序不同。

这道题是splay的基础题，也是很典型的不用AVl树做的题目（因为要保持平衡，AVl显得太过于繁复了）

对于CUT操作：

将当前树上第a-1个结点转到根结点，然后把第b+1个结点转到根结点的右儿子处，

于是root的右儿子的左儿子就是要删除区间的根，直接赋0就好

将c转到根结点，此时树被分成第1~第c的树A，和c的右子树B

将[a,b]这棵树接在第1~第c的右子结点上，

然后在新的树上找到最靠右的结点将c的右子树B结在它的右子结点上

对于Flip操作：

将当前树上第a-1个结点转到根结点，然后把第b+1个结点转到根结点的右儿子处，

于是root的右儿子的左儿子就是要要翻转的区间的根，

懒标记给此时root的右儿子的左儿子就好，

然后注意在所有的向下操作中都要pushdown()，即懒标记下传（这地方线段树的题体现得尤为明显）

写在代码前面：

为了防止出现一些不必要的诡异事件，我在树中加了两个虚拟结点（一个是-1，一个是N+1）方便调试

所以在调用找排名的函数时会将排名全部+1变成a和b+2，这并不是与我之前的分析相违背的

还有，如果直接建树会出现一条链，导致不够优化，所以此处我二分建树了。

呃呃。。它确实是跑得死慢死慢的。。。

20193787

2017-03-21 20:49:22

Accepted

3487

873MS

8624K

3521 B

G++

#include
#include
using namespace std;
const int Max=300000;
int N,M,root,cnt;
bool flag;
int ch[Max+5][2],fa[Max+5],fli[Max+5],num[Max+5],sz[Max+5];
void getint(int &num){
    char c;int flag=1;num=0;
    while((c=getchar())<'0'||c>'9')if(c=='-')flag=-1;
    while(c>='0'&&c<='9'){num=num*10+c-48;c=getchar();}
    num*=flag;
}
  
void change(int r){
    if(!r)return ;
    fli[r]^=1;
}
void pushup(int r){
    if(!r)return ;
    sz[r]=sz[ch[r][0]]+sz[ch[r][1]]+1;
}
void pushdown(int r){
    if(r&&fli[r]){
        swap(ch[r][0],ch[r][1]);
        change(ch[r][0]);
        change(ch[r][1]);
        change(r);
    }
}
void rotato(int x){
    if(!x||!fa[x])return ;
    int y=fa[x],z=fa[y];
	pushdown(y),pushdown(x);
    bool flag=(ch[y][1]==x);
    ch[y][flag]=ch[x][!flag];
    if(ch[x][!flag])    fa[ch[x][!flag]]=y;
    ch[x][!flag]=y,fa[y]=x;
    fa[x]=z;
    if(z)   ch[z][ch[z][1]==y]=x;
	pushup(y),pushup(x);
}
void splay(int x,int goal){
    for(int y; (y=fa[x])!=goal; rotato(x)){
        int z=fa[y];
        if(z!=goal){
            if((ch[y][1]==x)==(ch[z][1]==y))
                rotato(y);
            else rotato(x);
        }
    }
    if(goal==0) root=x;
}
void newtr(int &r,int f,int val){
    r=++cnt;
    fa[r]=f,sz[r]=1;
    num[r]=val;
    ch[r][0]=ch[r][1]=fli[r]=0;
}
void build(int &rt,int l,int r,int f){
    if(l>r)return ;
    int mid=(l+r)>>1;
    newtr(rt,f,mid);
    build(ch[rt][0],l,mid-1,rt);
    build(ch[rt][1],mid+1,r,rt);
    pushup(rt);
}
void Tr_build(){
    root=cnt=ch[0][0]=ch[0][1]=sz[0]=fa[0]=fli[0]=0;
    newtr(root,0,-1);
    newtr(ch[root][1],root,N+1);
    build(ch[ch[root][1]][0],1,N,ch[root][1]);
    pushup(ch[root][1]);
    pushup(root);
}
void Tr_merge(int root1,int root2){
    ch[root1][1]=root2;
    fa[root2]=root1;
}
int find_rank(int r,int rank){
    pushdown(r);
    int t=sz[ch[r][0]]+1;
    if(t==rank)return r;
    if(t>rank)   return find_rank(ch[r][0],rank);
    return find_rank(ch[r][1],rank-t);
}
int find_max(int r){
    pushdown(r);
    while(ch[r][1])
        r=ch[r][1],pushdown(r);
    return r;
}
void CUT(int a,int b,int c){
	int x=find_rank(root,a),z;
    splay(x,0);
	x=find_rank(root,b+2);
    splay(x,root);
    int root1=ch[ch[root][1]][0];
    ch[ch[root][1]][0]=0;
    pushup(ch[root][1]);
    pushup(root);
	z=find_rank(root,c+1);
    splay(z,0);
    int root2=ch[root][1];
    Tr_merge(root,root1);
    pushup(root);
    splay(find_max(root),0);
    Tr_merge(root,root2);
    pushup(root);
}
void Flip(int a,int b){
    splay(find_rank(root,a),0);
    splay(find_rank(root,b+2),root);
    change(ch[ch[root][1]][0]);
}
void print(int r){
    if(!ch[r][0]&&!ch[r][1]){
		if(num[r]!=-1&&num[r]!=N+1){
			if(flag)putchar(32);
			printf("%d",num[r]);
			flag=1;
		}
		return ;
    }
	pushdown(r);
    if(ch[r][0])print(ch[r][0]);
	if(num[r]!=-1&&num[r]!=N+1){
		if(flag)putchar(32);
		printf("%d",num[r]);flag=1;
	}
    if(ch[r][1])print(ch[r][1]);
}
int main(){
    char ord[5];
    int a,b,c;
    while(~scanf("%d%d",&N,&M)&&N!=-1&&M!=-1){
        Tr_build();
        for(int i=1; i<=M; ++i){
            scanf("%s",ord);
            getint(a),getint(b);
            if(ord[0]=='C'){
                getint(c);
                CUT(a,b,c);
            }
            else Flip(a,b);
        }
        flag=0;
        print(root);
		putchar(10);
    }
    return 0;
}