[bzoj3223]文艺平衡树

Description

 

您需要写一种数据结构（可参考题目标题），来维护一个有序数列，其中需要提供以下操作：翻转一个区间，例如原有序序列是5 4 3 2 1，翻转区间是[2,4]的话，结果是5 2 3 4 1 

Input

第一行为n,m n表示初始序列有n个数，这个序列依次是(1,2……n-1,n)  m表示翻转操作次数
接下来m行每行两个数[l,r] 数据保证 1<=l<=r<=n 

 

Output

 

输出一行n个数字，表示原始序列经过m次变换后的结果 

 

Sample Input

5 3

1 3

1 3

1 4

Sample Output

4 3 2 1 5

HINT

N,M<=100000

Source

平衡树

 

做完了维修数列之后，这就是个大水题

按题意反转序列，最后更新所有节点，输出中序遍历即可。。。。。。

 

// It is made by XZZ
#include<cstdio>
#include<algorithm>
#define Fname "sph"
using namespace std;
#define rep(a,b,c) for(rg int a=b;a<=c;a++)
#define drep(a,b,c) for(rg int a=b;a>=c;a--)
#define erep(a,b) for(rg int a=fir[b];a;a=nxt[a])
#define il inline
#define rg register
#define vd void
#define Now t[now]
#define pr pair<int,int>
#define mp make_pair
typedef long long ll;
il int gi(){
    rg int x=0,f=1;rg char ch=getchar();
    while(ch<'0'||ch>'9')f=ch=='-'?-1:f,ch=getchar();
    while(ch>='0'&&ch<='9')x=x*10+ch-'0',ch=getchar();
    return x*f;
}
ll seed=19260817;
int root;
const int maxn=101010;
il int Rand(){return (int)(seed=seed*48271%2147483647);}
struct node{int data,size,ls,rs,rand;bool rev;}t[maxn];
int newnode(int i){t[i]=(node){i,1,0,0,Rand(),0};return i;}
il vd down(int now){
    if(now&&Now.rev)swap(Now.ls,Now.rs),Now.rev=0,t[Now.ls].rev^=1,t[Now.rs].rev^=1;
}
il vd reset(int now){
    down(Now.ls),down(Now.rs);
    Now.size=t[Now.ls].size+t[Now.rs].size+1;
}
il int build(int n){
    int stack[maxn],top=0,last;
    rep(i,1,n){
    int now=newnode(i);last=0;
    while(top&&t[stack[top]].rand>Now.rand)reset(stack[top]),last=stack[top--];
    if(top)t[stack[top]].rs=now;Now.ls=last,stack[++top]=now;
    }
    while(top)reset(stack[top--]);
    return stack[1];
}
il int merge(int a,int b){
    if(!a||!b)return a|b;
    if(t[a].rand<t[b].rand){down(a),t[a].rs=merge(t[a].rs,b),reset(a);return a;}
    else{down(b),t[b].ls=merge(a,t[b].ls),reset(b);return b;}
}
il pr split(int now,int num){
    if(!now)return mp(0,0);
    down(now);
    int ls=Now.ls,rs=Now.rs;
    if(num==t[Now.ls].size){Now.ls=0,reset(now);return mp(ls,now);}
    if(num==t[Now.ls].size+1){Now.rs=0,reset(now);return mp(now,rs);}
    if(num<t[Now.ls].size){
    pr T=split(Now.ls,num);
    Now.ls=T.second,reset(now);
    return mp(T.first,now);
    }else{
    pr T=split(Now.rs,num-t[Now.ls].size-1);
    Now.rs=T.first,reset(now);
    return mp(now,T.second);
    }
}
il vd dfs(int now){
    if(now)down(Now.ls),down(Now.rs),dfs(Now.ls),printf("%d ",Now.data),dfs(Now.rs);
}
int main(){
    freopen(Fname".in","r",stdin);
    freopen(Fname".out","w",stdout);
    int n=gi(),m=gi();
    root=build(n);
    while(m--){
    int pos=gi(),tot=gi()-pos+1;
    pr T=split(root,pos-1),TT=split(T.second,tot);
    t[TT.first].rev^=1;
    root=merge(T.first,merge(TT.first,TT.second));
    }
    down(root),dfs(root);
    return 0;
}