BZOJ 3689: 异或之 可持久化trie+堆

和超级钢琴几乎是同一道题吧...

code:  

#include <bits/stdc++.h>    
#define N 200006   
#define ll long long 
#define setIO(s) freopen(s".in","r",stdin)  , freopen(s".out","w",stdout)   
using namespace std;
char buf[100000],*p1,*p2;
#define nc() (p1==p2&&(p2=(p1=buf)+fread(buf,1,100000,stdin),p1==p2)?EOF:*p1++)
int rd()
{
    int x=0; char s=nc();
    while(s<'0') s=nc();
    while(s>='0') x=(((x<<2)+x)<<1)+s-'0',s=nc();
    return x;
}                   
namespace trie 
{      
    int tot; 
    int cnt[N*30],ch[N*30][2];     
    int newnode() { return ++tot; }     
    void Insert(int pre,int &x,int v) 
    { 
        int now=x=newnode(),i; 
        for(i=30;i>=0;--i) 
        {
            int o=((v>>i)&1);     
            ch[now][o^1]=ch[pre][o^1]; 
            ch[now][o]=newnode();  
            pre=ch[pre][o]; 
            now=ch[now][o];   
            cnt[now]=cnt[pre]+1;  
        }
    }   
    int query(int x,int y,int z)      
    {
        int re=0,i;  
        for(i=30;i>=0;--i) 
        {
            int o=((z>>i)&1);   
            if(ch[x][o]<ch[y][o])   x=ch[x][o],y=ch[y][o];   
            else re+=(1<<i),x=ch[x][o^1],y=ch[y][o^1];             
        }
        return re;  
    }         
}; 
struct node 
{
    int o,l,r,val,pos;      
    node(int o=0,int l=0,int r=0,int val=0,int pos=0):o(o),l(l),r(r),val(val),pos(pos){}   
    bool operator<(node b) const 
    {
        return b.val<val; 
    }
}; 
priority_queue<node>q;     
int A[N],rt[N],ar[N],id[N];      
set<int>S[N];  
set<int>::iterator it;   
int main() 
{ 
    // setIO("input");  
    int i,j,n,k,ou=0; 
    n=rd(),k=rd();            
    for(i=1;i<=n;++i)   
    {
        A[i]=rd();              
        id[i]=ar[i]=A[i];                 
        trie::Insert(rt[i-1],rt[i],A[i]);       
    }           
    sort(ar+1,ar+1+n);   
    for(i=1;i<=n;++i)    id[i]=lower_bound(ar+1,ar+1+n,id[i])-ar;          
    for(i=1;i<=n;++i)    S[id[i]].insert(i);   
    for(i=1;i<n;++i) 
    {   
        int l=i+1,r=n;  
        int tmp=trie::query(rt[l-1],rt[r],A[i]);              
        int idx=lower_bound(ar+1,ar+1+n,A[i]^tmp)-ar;              
        int pos=*S[idx].lower_bound(l);         
        q.push(node(i,l,r,tmp,pos));               
    }    
    while(ou<k) 
    {
        node e=q.top(); q.pop();   
        printf("%d ",e.val),++ou;              
        int pos=e.pos;                        
        if(pos!=e.l)     
        {   
            int tmp=trie::query(rt[e.l-1],rt[pos-1],A[e.o]);   
            int idx=lower_bound(ar+1,ar+1+n,A[e.o]^tmp)-ar;           
            int t=*S[idx].lower_bound(e.l); 
            q.push(node(e.o,e.l,pos-1,tmp,t));   
        }
        if(pos!=e.r)  
        {
            int tmp=trie::query(rt[pos],rt[e.r],A[e.o]);  
            int idx=lower_bound(ar+1,ar+1+n,A[e.o]^tmp)-ar;       
            int t=*S[idx].lower_bound(pos+1);                  
            q.push(node(e.o,pos+1,e.r,tmp,t));      
        }
    }
    return 0; 
}

  

 

posted @ 2019-11-26 20:15  EM-LGH  阅读(132)  评论(0编辑  收藏  举报