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

51 2 3 4 532 2 4 21 3 62 2 4 2

 

Sample Output

34

 

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较紧！貌似用树状数组套主席树会T

需要线段树套treap！

代码如下：

//#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 3.1415926535897932384626433832795const double e = exp(1.0);#define INF 0x3f3f3f3f//#define INF 1e18//typedef long long LL;//typedef __int64 LL;#define ONLINE_JUDGE#ifndef ONLINE_JUDGEfreopen("in.txt", "r", stdin);freopen("out.txt", "w", stdout);#endif#define N 600010#define M 100010struct 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\n",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;}