HDU 5412 CRB and Queries(区间第K大 树套树 按值建树)

题目链接：http://acm.hdu.edu.cn/showproblem.php?

pid=5412

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

题意：

给出一串数字。然后给出两种操作：

1：1 L V  操作一：把下标为L的点的值替换为 V

2：2 L R K  操作二：在[L, R]区间求第K大！

PS:

这题好像时间卡的比較紧！

貌似用树状数组套主席树会T

须要线段树套treap。

代码例如以下：

//#pragma warning (disable:4786)
//#pragma comment(linker,"/STACK:102400000,102400000")  //手动扩栈
//#include <bits/stdc++.h>
#include <cstdio>
#include <cmath>
#include <cstring>
#include <string>
#include <cstdlib>
#include <climits>
#include <ctype.h>
#include <queue>
#include <stack>
#include <vector>
#include <utility>
#include <deque>
#include <set>
#include <map>
#include <iostream>
#include <algorithm>
using namespace std;
const double eps = 1e-9;
const double PI = acos(-1.00);
//#define PI 3.1415926535897932384626433832795
const double e = exp(1.0);
#define INF 0x3f3f3f3f
//#define INF 1e18
//typedef long long LL;
//typedef __int64 LL;
#define ONLINE_JUDGE
#ifndef ONLINE_JUDGE
freopen("in.txt", "r", stdin);
freopen("out.txt", "w", stdout);
#endif


#define N 600010
#define M 100010
struct treap
{
    int key,wht,count,sz,ch[2];
} tp[N*15];
int tree[N<<1];
int nodecount,root;
int IDX(int l,int r)
{
    return l+r | l!=r;
}
void init()
{
    tp[0].sz=0;
    tp[0].wht=-INF;
    nodecount=0;
    root=0;
}
void update(int x)
{
    tp[x].sz=tp[tp[x].ch[0]].sz+tp[x].count+tp[tp[x].ch[1]].sz;
}
void rotate(int &x,int t)
{
    int y=tp[x].ch[t];
    tp[x].ch[t]=tp[y].ch[!t];
    tp[y].ch[!t]=x;
    update(x);
    update(y);
    x=y;
}
void insert(int &x,int t)
{
    if(! x)
    {
        x=++nodecount;
        tp[x].key=t;
        tp[x].wht=rand();
        tp[x].count=1;
        tp[x].ch[0]=tp[x].ch[1]=0;
    }
    else if(tp[x].key==t)  tp[x].count++;
    else
    {
        int k=tp[x].key<t;
        insert(tp[x].ch[k],t);
        if(tp[x].wht<tp[tp[x].ch[k]].wht) rotate(x,k);
    }
    update(x);
}
void erase(int &x,int t)
{
    if(tp[x].key==t)
    {
        if(tp[x].count==1)
        {
            if(! tp[x].ch[0] && ! tp[x].ch[1])
            {
                x=0;
                return;
            }
            rotate(x,tp[tp[x].ch[0]].wht<tp[tp[x].ch[1]].wht);
            erase(x,t);
        }
        else tp[x].count--;
    }
    else erase(tp[x].ch[tp[x].key<t],t);
    update(x);
}
int select(int x,int t)
{
    if(! x) return 0;
    if(tp[x].key>t) return select(tp[x].ch[0],t);
    return tp[x].count+tp[tp[x].ch[0]].sz+select(tp[x].ch[1],t);
}
int a[N],b[N],ord[M][5],lb;
int n,m,tt;
int search(int x)
{
    int l=1,r=b[0],mid;
    while (l<=r)
    {
        mid=(l+r)>>1;
        if(b[mid]==x) return mid;
        if(b[mid]<x) l=mid+1;
        else r=mid-1;
    }
}
void treeinsert(int l,int r,int i,int x)
{
    insert(tree[IDX(l,r)],x);
    if(l==r) return;
    int m=(l+r)>>1;
    if(i<=m) treeinsert(l,m,i,x);
    else treeinsert(m+1,r,i,x);
}
void treedel(int l,int r,int i,int x)
{
    erase(tree[IDX(l,r)],x);
    if(l==r) return;
    int m=(l+r)>>1;
    if(i<=m) treedel(l,m,i,x);
    else treedel(m+1,r,i,x);
}
int query(int l,int r,int x,int y,int k)
{
    if(l==r) return l;
    int m=(l+r)>>1;
    int ans=select(tree[IDX(l,m)],y)-select(tree[IDX(l,m)],x);
    if(ans>=k) return query(l,m,x,y,k);
    return query(m+1,r,x,y,k-ans);
}
int main ()
{
    while (~scanf("%d",&n))
    {
        b[0]=1;
        lb=0;
        memset(tree,0,sizeof(tree));
        init();
        for(int i=1; i<=n; i++)
        {
            scanf("%d",&a[i]);
            b[++lb]=a[i];
        }
        scanf("%d",&m);
        for(int i=1; i<=m; i++)
        {
            int op;
            int x,y,c;
            scanf("%d",&op);
            if(op == 2)
            {
                scanf("%d %d %d",&x,&y,&c);
                ord[i][1]=1;
                ord[i][2]=x;
                ord[i][3]=y;
                ord[i][4]=c;
            }
            else
            {
                scanf("%d %d",&x,&y);
                ord[i][1]=2;
                ord[i][2]=x;
                ord[i][3]=y;
                b[++lb]=y;
            }
        }
        sort(b+1,b+1+lb);
        for(int i=1; i<=lb; i++)
            if(b[i]!=b[b[0]]) b[++b[0]]=b[i];
        for(int i=1; i<=n; i++)
        {
            a[i]=search(a[i]);
            treeinsert(1,b[0],a[i],i);
        }
        for(int i=1; i<=m; i++)
        {
            if(ord[i][1]==1)
                printf("%d
",b[query(1,b[0],ord[i][2]-1,ord[i][3],ord[i][4])]);
            else
            {
                treedel(1,b[0],a[ord[i][2]],ord[i][2]);
                a[ord[i][2]]=search(ord[i][3]);
                treeinsert(1,b[0],a[ord[i][2]],ord[i][2]);
            }
        }
    }
    return 0;
}