树套树三题 题解

1.COGS 1534 [NEERC 2004]K小数

其实是主席树裸题……

(其实这题数据非常水……从O(nlogn)的主席树到O(nlog³n)的树套树+二分到O(nsqrt(n)log²n)的分块套二分套二分到O(n²)的暴力都能过……)

鉴于这就是动态排名系统的静态版，就不说了，贴代码：

线段树套平衡树：

#include<cstdio>
#include<cstring>
#include<algorithm>
using namespace std;
const int maxn=100010;
struct node{
    int data,size;
    node *lc,*rc,*prt;
    node(int d=0):data(d),size(1),lc(NULL),rc(NULL),prt(NULL){}
    void refresh(){
        size=1;
        if(lc)size+=lc->size;
        if(rc)size+=rc->size;
    }
}*root[maxn<<2]={NULL};
void build(int,int,int);
void query(int,int,int);
void insert(node*,int);
void copy(int,node*);
int rank(int,int);
void splay(node*,node*,int);
void lrot(node*,int);
void rrot(node*,int);
int n,m,x,l,r,k,s,t,tmp,ans;
int main(){
#define MINE
#ifdef MINE
    freopen("kthnumber.in","r",stdin);
    freopen("kthnumber.out","w",stdout);
#endif
    scanf("%d%d",&n,&m);
    build(1,n,1);
    while(m--){
        scanf("%d%d%d",&s,&t,&k);
        l=-(int)(1e9+0.5);r=-l;
        while(l<=r){
            ans=(l+r)>>1;
            tmp=0;
            query(1,n,1);
            if(tmp<k)l=ans+1;
            else r=ans-1;
        }
        printf("%d\n",r);
    }
#ifndef MINE
    printf("\n--------------------DONE--------------------\n");
    for(;;);
#else
    fclose(stdin);
    fclose(stdout);
#endif
    return 0;
}
void build(int l,int r,int rt){
    if(l==r){
        int x;
        scanf("%d",&x);
        insert(new node(x),rt);
        return;
    }
    int mid=(l+r)>>1;
    build(l,mid,rt<<1);
    build(mid+1,r,rt<<1|1);
    copy(rt,root[rt<<1]);
    copy(rt,root[rt<<1|1]);
}
void query(int l,int r,int rt){
    if(s<=l&&t>=r){
        tmp+=rank(ans,rt);
        return;
    }
    int mid=(l+r)>>1;
    if(s<=mid)query(l,mid,rt<<1);
    if(t>mid)query(mid+1,r,rt<<1|1);
}
void copy(int rt,node *x){
    if(!x)return;
    insert(new node(x->data),rt);
    copy(rt,x->lc);
    copy(rt,x->rc);
}
void insert(node *x,int i){
    if(!root[i]){
        root[i]=x;
        return;
    }
    node *rt=root[i];
    for(;;){
        if(x->data<rt->data){
            if(rt->lc)rt=rt->lc;
            else{
                rt->lc=x;
                break;
            }
        }
        else{
            if(rt->rc)rt=rt->rc;
            else{
                rt->rc=x;
                break;
            }
        }
    }
    x->prt=rt;
    while(rt){
        rt->refresh();
        rt=rt->prt;
    }
    splay(x,NULL,i);
}
int rank(int x,int i){
    node *rt=root[i],*y=rt;
    int ans=0;
    while(rt){
        y=rt;
        if(x<=rt->data)rt=rt->lc;
        else{
            if(rt->lc)ans+=rt->lc->size;
            ans++;
            rt=rt->rc;
        }
    }
    splay(y,NULL,i);
    return ans;
}
void splay(node *x,node *tar,int i){
    if(!x)return;
    for(node *rt=x->prt;rt!=tar;rt=x->prt){
        if(rt->prt==tar){
            if(x==rt->lc)rrot(rt,i);
            else lrot(rt,i);
            break;
        }
        if(rt==rt->prt->lc){
            if(x==rt->lc)rrot(rt,i);
            else lrot(rt,i);
            rrot(x->prt,i);
        }
        else{
            if(x==rt->rc)lrot(rt,i);
            else rrot(rt,i);
            lrot(x->prt,i);
        }
    }
}
void lrot(node *x,int i){
    node *y=x->rc;
    if(x->prt){
        if(x==x->prt->lc)x->prt->lc=y;
        else x->prt->rc=y;
    }
    else root[i]=y;
    y->prt=x->prt;
    x->rc=y->lc;
    if(y->lc)y->lc->prt=x;
    y->lc=x;
    x->prt=y;
    x->refresh();
    y->refresh();
}
void rrot(node *x,int i){
    node *y=x->lc;
    if(x->prt){
        if(x==x->prt->lc)x->prt->lc=y;
        else x->prt->rc=y;
    }
    else root[i]=y;
    y->prt=x->prt;
    x->lc=y->rc;
    if(y->rc)y->rc->prt=x;
    y->rc=x;
    x->prt=y;
    x->refresh();
    y->refresh();
}

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