HDU 5412——CRB and Queries——————【线段树套Treap（并没有AC）】

CRB and Queries

Time Limit: 12000/6000 MS (Java/Others)    Memory Limit: 131072/131072 K (Java/Others)
Total Submission(s): 1602    Accepted Submission(s): 409

Problem Description

There are N boys in CodeLand.
Boy i has his coding skill Ai.
CRB wants to know who has the suitable coding skill.
So you should treat the following two types of queries.
Query 1: 1 l v
The coding skill of Boy l has changed to v.
Query 2: 2 l r k
This is a report query which asks the k-th smallest value of coding skill between Boy l and Boy r(both inclusive).

 

Input

There are multiple test cases. 
The first line contains a single integer N.
Next line contains N space separated integers A1, A2, …, AN, where Ai denotes initial coding skill of Boy i.
Next line contains a single integer Q representing the number of queries.
Next Q lines contain queries which can be any of the two types.
1 ≤ N, Q ≤ 105
1 ≤ Ai, v ≤ 109
1 ≤ l ≤ r ≤ N
1 ≤ k ≤ r – l + 1

 

Output

For each query of type 2, output a single integer corresponding to the answer in a single line.

 

Sample Input

5

1 2 3 4 5

3

2 2 4 2

1 3 6

2 2 4 2

 

Sample Output

3

4

 

Author

KUT（DPRK）

 

Source

2015 Multi-University Training Contest 10

 

题目大意：给你一个数列有n个数，q次询问。询问1可以把某个位置改为值b。询问2然后问L - R之间第k大的数是多少。简单来说，就是带修改的动态区间第k大问题。

解题思路：用线段树维护Ci在去重后的所有可能出现在数列中的数当中的大小排名，用Treap维护Ci在数列中的位置关系。其实我觉得这才是内涵。然后对于查询区间L - R的第k大。那么如果在线段树左儿子代表的Treap树中在R位置之前的个数减去左儿子代表的Treap树中的(L-1)之前的个数大于k，那么就可以在左儿子中找位置。

#include<stdio.h>
#include<string.h>
#include<time.h>
#include<algorithm>
using namespace std;
#define mid (L+R)/2
#define lson rt*2,L,mid
#define rson rt*2+1,mid+1,R
const int maxn=1e6;
struct Treap{
    int rk;
    int coun;
    int sz;
    int v;
    Treap *ch[2];
    Treap() {
        coun=0;
        sz=0;
        rk=-maxn;   //
        v=0;
    }
    int cmp(int x) const {
        if(x==v){
            return -1;
        }
        return x<v ? 0:1;
    }
};
Treap *seg[maxn*8];
Treap *null;
int oper[maxn][4];
int a[maxn],b[maxn*3];
int discr(int l,int r,int key){
    while(l<=r){
        int md=(l+r)/2;
        if(key<b[md]){
            r=md-1;
        }else if(key > b[md]){
            l=md+1;
        }else{
            return md;
        }
    }
    return -1;
}
void update(Treap * &o){
    o->sz = o->ch[0]->sz + o->coun + o->ch[1]->sz;
}
void rotate(Treap * &o,int d){
    Treap *k=o->ch[d^1]; o->ch[d^1]= k->ch[d]; k->ch[d]=o;
    update(o); update(k);  o=k;
}
void insert_tp(Treap * &o,int x){
    if(o==null){
        o = new Treap();
        o->ch[0]=o->ch[1]=null;
        o->v = x; o->coun = 1; o->rk = rand();
    }else{
        int d=o->cmp(x);
        insert_tp(o->ch[d],x);
        if(o->ch[d]->rk > o->rk) rotate(o,d^1);
    }
    update(o);
}
void insert_seg(int rt,int L,int R,int pos,int x){
    insert_tp(seg[rt],x);
    if(L==R)
        return ;
    if(pos<=mid){
        insert_seg(lson,pos,x);
    }else{
        insert_seg(rson,pos,x);
    }
}
void free_tp(Treap *&o){
    if(o->ch[0]==null&&o->ch[1]==null){
        free(o);
        return ;
    }
    if(o->ch[0]!=null){
        free_tp(o->ch[0]);
    }
    if(o->ch[1]!=null){
        free_tp(o->ch[1]);
    }
    free(o);
}
void clean(int rt,int L,int R){
    if(L==R){
        free_tp(seg[rt]);
        return ;
    }
    clean(lson);
    clean(rson);
    free_tp(seg[rt]);
}
void del_tp(Treap * &o,int x){
    int d=o->cmp(x);
    if(d==-1){
        if(o->ch[0]==null&&o->ch[1]==null){     //
            Treap *pt=o;
            o = null;
            free(pt);
        }
        else if(o->ch[0]==null){
            Treap *pt=o;
            o=o->ch[1];
            free(pt);
        }else if(o->ch[1]==null){
            Treap *pt=o;
            o= o->ch[0];
            free(pt);
        }else{
            int d2=( o->ch[0]->rk >o->ch[1]->rk ? 1:0 );
            rotate(o,d2);
            del_tp(o->ch[d2],x);
        }
    }else{
        del_tp(o->ch[d],x);
    }
    update(o);
}
void del_seg(int rt,int L,int R,int pos,int x){
    del_tp(seg[rt],x);
    if(L==R)
        return ;
    if(pos<=mid){
        del_seg(lson,pos,x);
    }else{
        del_seg(rson,pos,x);
    }
}
int select(Treap *o,int x){
    if(o==null){
        return 0;
    }
    if(o->v > x) return select(o->ch[0],x);
    return o->ch[0]->sz + o->coun +select(o->ch[1],x);
}
int query(int rt,int L,int R,int x,int y,int k){
    if(L==R) return L;
    int ans=select(seg[rt*2],y)-select(seg[rt*2],x);
    if(ans>=k)
        return query(lson,x,y,k);
    else return query(rson,x,y,k-ans);
}
void init(){
    null =new Treap();
    null->ch[0]=null->ch[1]=null;
    null->sz = null->coun=0;
    for(int i=0;i<=maxn*4-1;i++){
        seg[i] = null;
    }
}
int main(){
//    freopen("1007.in","r",stdin);
//    freopen("OUT.txt","w",stdout);
    int n,Q,typ,x,y,z,nn,mm;
    while(scanf("%d",&n)!=EOF){
        init();
        mm=0;
        for(int i=0;i<n;i++){
            scanf("%d",&a[i]);
            b[mm++]=a[i];
        }
        scanf("%d",&Q);
        for(int i=0;i<Q;i++){
            scanf("%d",&typ);
            oper[i][0]=typ;
            if(typ==2){
                scanf("%d%d%d",&x,&y,&z);
                oper[i][1]=x;
                oper[i][2]=y;
                oper[i][3]=z;
            }else{
                scanf("%d%d",&x,&z);
                oper[i][1]=x;
                oper[i][2]=z;
                b[mm++]=z;
            }
        }
        sort(b,b+mm);
        nn=1;
        for(int i=1;i<mm;i++){
            if(b[i]!=b[i-1]){
                b[nn++]=b[i];
            }
        }
        for(int i=0;i<n;i++){
            a[i]= discr(0,nn-1,a[i])+1;
            insert_seg(1,1,nn,a[i],i+1);
        }
        for(int i=0;i<Q;i++){
            if(oper[i][0]==1){
                int kk=0;
                del_seg(1,1,nn,a[oper[i][1]-1],oper[i][1]);
                kk=discr(0,nn-1,oper[i][2]);
                a[oper[i][1]-1]=kk+1;
                insert_seg(1,1,nn,a[oper[i][1]-1],oper[i][1]);
            }else{
                int tmp=query(1,1,nn,oper[i][1]-1,oper[i][2],oper[i][3])-1;
                printf("%d\n",b[tmp]);
            }
        }
        clean(1,1,nn);
    }
    return 0;
}

/*

5
1 1 1 1 1
10
1 2 3
2 1 4 3
1 5 6
1 4 2
1 5 3
2 1 3 3
2 1 4 4
2 2 5 2
2 2 5 3
2 2 5 4

*/