poj2104 第k小数

Posted 2020-10-15

Description

　　You are working for Macrohard company in data structures department. After failing your previous task about key insertion you were asked to write a new data structure that would be able to return quickly k-th order statistics in the array segment. 
　　That is, given an array a[1...n] of different integer numbers, your program must answer a series of questions Q(i, j, k) in the form: "What would be the k-th number in a[i...j] segment, if this segment was sorted?" 
　　For example, consider the array a = (1, 5, 2, 6, 3, 7, 4). Let the question be Q(2, 5, 3). The segment a[2...5] is (5, 2, 6, 3). If we sort this segment, we get (2, 3, 5, 6), the third number is 5, and therefore the answer to the question is 5.

Input

　　The first line of the input file contains n --- the size of the array, and m --- the number of questions to answer (1 <= n <= 100 000, 1 <= m <= 5 000). 
　　The second line contains n different integer numbers not exceeding 10⁹ by their absolute values --- the array for which the answers should be given. 
　　The following m lines contain question descriptions, each description consists of three numbers: i, j, and k (1 <= i <= j <= n, 1 <= k <= j - i + 1) and represents the question Q(i, j, k).

Output

　　For each question output the answer to it --- the k-th number in sorted a[i...j] segment.

Sample Input

　　7 3
　　1 5 2 6 3 7 4
　　2 5 3
　　4 4 1
　　1 7 3

Sample Output

　　5
　　6
　　3

题解：　

　　可持久化线段树，又叫主席树。作为模板题我就多说说吧。

　　可持久化的意思就是可以随时查询线段树维护过程中任何状态。主席树是权值线段树，就是每个值代表一个点，在定义域上建立线段树，[1,1],[1,2],[1,3]...[1,n]这n颗分别维护区间中的信息。这题我们就让主席树变成前缀线段树，然后查询的话就是把第R颗与第L-1颗的差做出来去查询即可

　　主席树还是动态开点的线段树。（具体看代码）

 1 #include<iostream>
 2 #include<cstdlib>
 3 #include<cstring>
 4 #include<cstdio>
 5 #include<algorithm>
 6 using namespace std;
 7 int rt[100005],a[100005],c[100005],s[100005];
 8 struct node{
 9     int l,r,sz,lc,rc;
10 }    tr[2000005];
11 int cnt,tot;
12 inline void update(int x,int y,int p){
13     tr[x]=tr[y];
14     tr[x].sz++;
15     int l=tr[x].l,r=tr[x].r;
16     if(l==r)    return;
17     int mid=(l+r)>>1;
18     if(p<=mid){
19         tr[x].lc=++cnt;
20         update(tr[x].lc,tr[y].lc,p);
21     }
22     else{
23         tr[x].rc=++cnt;
24         update(tr[x].rc,tr[y].rc,p);
25     }
26 }
27 inline void build(int rt,int l,int r)
28 {
29     tr[rt].l=l,tr[rt].r=r;
30     if(l==r)    return;
31     int mid=(l+r)>>1;
32     tr[rt].lc=++cnt;    build(tr[rt].lc,l,mid);
33     tr[rt].rc=++cnt;    build(tr[rt].rc,mid+1,r);
34 }
35 inline int query(int x,int y,int p)
36 {
37     int l=tr[x].l,r=tr[x].r;
38     if(l==r)    return l;
39     if(p<=tr[tr[x].lc].sz-tr[tr[y].lc].sz){
40         return query(tr[x].lc,tr[y].lc,p);
41     }
42     else{
43         return query(tr[x].rc,tr[y].rc,p-(tr[tr[x].lc].sz-tr[tr[y].lc].sz));
44     }
45 }
46 int main(){
47     int n,q,L,R,k;
48     scanf("%d%d",&n,&q);
49     for(int i=1;i<=n;i++){
50         scanf("%d",&a[i]);
51         c[i]=a[i];
52     }
53     sort(c+1,c+n+1);
54     for(int i=1;i<=n;i++){
55         if(c[i]!=c[i-1])    s[++tot]=c[i];
56     }
57     rt[0]=++cnt;
58     build(rt[0],1,tot);
59     for(int i=1;i<=n;i++){
60         int p=lower_bound(s+1,s+tot+1,a[i])-s;
61         rt[i]=++cnt;
62         update(rt[i],rt[i-1],p);
63     }
64     while(q--){
65         scanf("%d%d%d",&L,&R,&k);
66         printf("%d\n",s[query(rt[R],rt[L-1],k)]);
67     }
68     return 0;
69 }