bzoj 3223 Tyvj 1729 文艺平衡树 [Splay]

文艺平衡树

Time Limit: 10 Sec
Memory Limit: 128 MB
Submit: 3386 Solved: 1916

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

---

Splay细节太多。。建树的时候 l>r 要return！！
然后就是注意update！！
以及最后遍历的时候注意标记下放！！

#include<iostream>#include<cstdio>#include<cstring>#include<cstdlib>#include<cmath>#include<vector>#include<queue>#include<stack>#include<map>#include<set>#include<string>#include<iomanip>#include<ctime>#include<climits>#include<cctype>#include<algorithm>#ifdef WIN32#define AUTO "%I64d"#else#define AUTO "%lld"#endifusing namespace std;#define smax(x,tmp) x=max((x),(tmp))#define smin(x,tmp) x=min((x),(tmp))#define maxx(x1,x2,x3) max(max(x1,x2),x3)#define minn(x1,x2,x3) min(min(x1,x2),x3)const int INF=0x3f3f3f3f;const int maxn = 100005;struct Node{    int val;    int size;    int fa,ch[2];    bool rev;}node[maxn];int root;#define val(x) node[x].val#define size(x) node[x].size#define fa(x) node[x].fa#define ch(x,d) node[x].ch[d]#define rev(x) node[x].revinline void update(int x){    int l=ch(x,0),r=ch(x,1);    size(x)=1;    if(l) size(x) += size(l);    if(r) size(x) += size(r);}void rotate(int x,int &to){    int y=fa(x),z=fa(y);    int l= ch(y,1)==x , r=l^1;    if(y==to) to = x;    else ch(z,ch(z,1)==y) = x;    fa(ch(x,r))=y; fa(y)=x; fa(x)=z;    ch(y,l)=ch(x,r); ch(x,r)=y;    update(y); update(x);}void splay(int x,int &to){    while(x^to)    {        int y=fa(x),z=fa(y);        if(y^to)            if(ch(y,0)==x ^ ch(z,0)==y) rotate(x,to);            else rotate(y,to);        rotate(x,to);    }}void pushdown(int x){    if(!rev(x)) return;    rev(x)^=1;    int l=ch(x,0),r=ch(x,1);    if(l) swap(ch(l,0),ch(l,1)),rev(l)^=1;    if(r) swap(ch(r,0),ch(r,1)),rev(r)^=1;}int find(int x,int k){    pushdown(x);    int lsize=size(ch(x,0));    if(lsize+1==k) return x;    if(k<=lsize) return find(ch(x,0),k);    else return find(ch(x,1),k-lsize-1);}int split(int s,int t) // original array{    int x=find(root,s),y=find(root,t+2);    splay(x,root); splay(y,ch(x,1));    return ch(y,0);}void reverse(int s,int t){    int rt = split(s,t);    swap(ch(rt,0),ch(rt,1));    rev(rt) ^= 1;}int n,m;int a[maxn];void build(int l,int r,int f){    if(l>r) return; // must !! or RE!!    int m=(l+r)>>1;    if(l^r) build(l,m-1,m),build(m+1,r,m);    fa(m)=f;    ch(f,m>f)=m; // !!    val(m)=a[m];    update(m); // !!}void print_in(int x){    if(!x) return;    pushdown(x); // must !! probably to be lazyed!    print_in(ch(x,0));    if(val(x)^INF) printf("%d ",val(x));    print_in(ch(x,1));}int main(){    freopen("splay.in","r",stdin);    freopen("splay.out","w",stdout);    scanf("%d%d",&n,&m);    a[1]=a[n+2]=INF;    for(int i=1;i<=n;i++) a[i+1]=i;    build(1,n+2,0);    root = (n+3)>>1;    for(int i=1;i<=m;i++)    {        int s,t;        scanf("%d%d",&s,&t);        reverse(s,t);        //print_in(root);cout<<endl;    }    print_in(root);}