Codeforces - 675D 可持久化Treap 树形操作

题意:模拟二叉树的构造过程,给出\(n\)个节点,每次从根插入,小于当前节点转到左儿子,否则右儿子,输出第\([2,n]\)个节点的父亲的权值

直接手动模拟会被链式结构T掉

网上找了下发现二叉树的性质是当前插入节点的父亲总是比它小的最大值或比它大的最小值(深度最深者为父亲)

既然这样就容易搞了,Treap分裂找出这两个值比较并维护即可(数据保证每个值只出现一次,更容易维护了)

93ms效率非常可观,可持久化Treap天下第一

#include<iostream>
#include<algorithm>
#include<cstdio>
#include<cstring>
#include<cstdlib>
#include<cmath>
#include<string>
#include<vector>
#include<stack>
#include<queue>
#include<set>
#include<map>
#define rep(i,j,k) for(register int i=j;i<=k;i++)
#define rrep(i,j,k) for(register int i=j;i>=k;i--)
#define erep(i,u) for(register int i=head[u];~i;i=nxt[i])
#define iin(a) scanf("%d",&a)
#define lin(a) scanf("%lld",&a)
#define din(a) scanf("%lf",&a)
#define s0(a) scanf("%s",a)
#define s1(a) scanf("%s",a+1)
#define print(a) printf("%lld",(ll)a)
#define enter putchar('\n')
#define blank putchar(' ')
#define println(a) printf("%lld\n",(ll)a)
#define IOS ios::sync_with_stdio(0)
using namespace std;
const int MAXN = 2e5+11;
const int INF = 0x3f3f3f3f;
const double EPS = 1e-7;
typedef long long ll;
const ll MOD = 1e9+7;
unsigned int SEED = 19260817;
ll read(){
    ll x=0,f=1;register char ch=getchar();
    while(ch<'0'||ch>'9'){if(ch=='-')f=-1;ch=getchar();}
    while(ch>='0'&&ch<='9'){x=x*10+ch-'0';ch=getchar();}
    return x*f;
}
inline int Rand(){
    SEED=SEED*1103515245+12345;
    return SEED/65536;
}
struct Treap{
    int son[MAXN][2],root,tot;
    int val[MAXN],fix[MAXN],size[MAXN];
    int dep[MAXN];
    #define lc son[o][0]
    #define rc son[o][1]
    void init(){
        root=0;
        son[0][0]=son[0][1]=0;
        val[0]=fix[0]=size[0]=0;
        dep[0]=0;
        tot=1;
    }
    int node(int v,int d){
        son[tot][0]=son[tot][1]=0;
        val[tot]=v; fix[tot]=Rand();
        size[tot]=1;dep[tot]=d;
        return tot++;
    }
    void pu(int o){
        size[o]=size[lc]+size[rc]+1;
    }
    void split(int o,int pivot,int &a,int &b){
        if(!o){
            a=b=0;
            return;
        }else if(val[o]>pivot){
            b=o;
            split(lc,pivot,a,lc);
            pu(o);
        }else{
            a=o;
            split(rc,pivot,rc,b);
            pu(o);
        }
    }
    int merge(int a,int b){
        if(!a) return b;
        if(!b) return a;
        if(fix[a]<fix[b]){
            son[a][1]=merge(son[a][1],b);
            pu(a);
            return a;
        }else{
            son[b][0]=merge(a,son[b][0]);
            pu(b);
            return b;
        }
    }
    void insert(int v,int d){
        int a,b,t=node(v,d);
        split(root,v,a,b);
        root=merge(merge(a,t),b);
    }
    int kth(int o,int k){
    	if(!o)return o;//
        while(1){
            if(k<=size[lc]){
                o=lc;
            }else if(k==size[lc]+1){
                return o;
            }else{
                k-=size[lc]+1;
                o=rc;
            }
        }
    }
    int insert(int v){
        int a,b;
        split(root,v-1,a,b);
        int x=kth(a,size[a]);
        int y=kth(b,1);
        bool flag=0;
        if(dep[x]<dep[y]) flag=1;
        root=merge(a,b);
        int d=(flag?dep[y]:dep[x])+1;
        insert(v,d);
        return flag?val[y]:val[x];
    }
}tp;
int n,m,a[MAXN],ans[MAXN];
int main(){
    while(cin>>n){
        tp.init();
        rep(i,1,n) a[i]=read();tp.insert(a[1],1);
        rep(i,2,n) ans[i]=tp.insert(a[i]);
        rep(i,2,n) printf("%d%c",ans[i],i==n?'\n':' ');
    }
    return 0;
}