BZOJ 3223 文艺平衡树

3223: Tyvj 1729 文艺平衡树

Time Limit: 10 Sec  Memory Limit: 128 MB
Submit: 5402  Solved: 3196
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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

正如sourse所说这是一道平衡树的题目，treep对于区间维护好像并不太擅长

我们用splay，splay改变序列是靠中序遍历的不同，区间反转维护lazy标记即可

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#include <bits/stdc++.h> #define ll long long #define inf 10000100 using namespace std; inline int read(){     int x=0;int f=1;char ch=getchar();     while(!isdigit(ch)) {if(ch=='-') f=-1;ch=getchar();}     while(isdigit(ch)) {x=x*10+ch-'0';ch=getchar();}     return x*f; } const int MAXN=1e6+10; namespace zhangenming{     int son[MAXN][2],n,m,siz[MAXN],rev[MAXN]={},fa[MAXN],a[MAXN],tot,key[MAXN],root=0;     inline void push_up(int x){         siz[x]=siz[son[x][1]]+siz[son[x][0]]+1;     }     inline int get(int x){         return x==son[fa[x]][1];     }     inline void push_down(int x){         if(!x||!rev[x]) return;         rev[son[x][1]]^=1;rev[son[x][0]]^=1;         swap(son[x][1],son[x][0]);         rev[x]=0;     }     inline void rote(int x){         push_down(x);push_down(fa[x]);         int oldl=fa[x];int wei=get(x);int oldf=fa[fa[x]];         son[oldl][wei]=son[x][wei^1];fa[son[oldl][wei]]=oldl;         fa[x]=oldf;fa[oldl]=x;son[x][wei^1]=oldl;         if(oldf){             son[oldf][son[oldf][1]==oldl]=x;         }         push_up(x);push_up(oldl);     }     inline int build(int leftt,int rightt,int root){         if(leftt>rightt) return 0;         int mid=(rightt+leftt)>>1;         int now=++tot;         rev[now]=0;fa[now]=root;key[now]=a[mid];         int lefttchild=build(leftt,mid-1,now);         int righttchild=build(mid+1,rightt,now);         son[now][0]=lefttchild;son[now][1]=righttchild;push_up(now);         return now;     }     inline void splay(int xx,int tal){         for(int f;(f=fa[xx])!=tal;rote(xx)){             if(fa[f]!=tal){                 rote(get(f)==get(xx)?f:xx);             }         }         if(tal==0) root=xx;     }     inline int find(int xx){         int now=root;         while(1){             push_down(now);             if(xx==0) return now;             if(xx<=siz[son[now][0]]){                 now=son[now][0];             }             else{                 xx-=siz[son[now][0]]+1;                 if(xx==0) return now;                 now=son[now][1];             }         }     }     void print(int root){         push_down(root);         if(son[root][0]) print(son[root][0]);         if(key[root]!=inf&&key[root]!=-inf) printf("%d ",key[root]);         if(son[root][1]) print(son[root][1]);     }     void init(){         n=read();m=read();         a[1]=inf;a[n+2]=-inf;         for(int i=1;i<=n;i++){             a[i+1]=i;         }         root=build(1,n+2,0);     }     inline void printt(){         for(int i=1;i<=n+2;i++){             cout<<son[i][0]<<' '<<son[i][1]<<endl;         }     }     void solve(){         while(m--){             int xx=read();int yy=read();             if(xx>=yy) continue;             xx=find(xx);yy=find(yy+2);             splay(xx,0);splay(yy,xx);             rev[son[son[root][1]][0]]^=1;         }         print(root);     } } int main(){     //freopen("All.in","r",stdin);     //freopen("a.out","w",stdout);     using namespace zhangenming;     init();     solve();     return 0; }