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).
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
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
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
*/
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