HDU 2665 durable segment tree for interval k large value (functional Segment Tree | | Chairman Tree)

http://acm.hdu.edu.cn/showproblem.php?pid=2665

Problem descriptiongive you a sequence and ask you the kth big number of a inteval.
Inputthe first line is the number of the test cases.
For each test case, the first line contain both integer n and m (n, M <= 100000), indicates the number of integers in th E sequence and the number of the quaere.
The second line contains n integers, describe the sequence.
Each of following m lines contains three integers s, T, K.
[s, T] indicates the interval and K indicates the kth big number in interval [s, T]
Outputfor each test case, output m lines. Each line contains the kth big number.
Sample Input

Sample Output

2

/**HDU 2665 durable segment tree for interval k large value (functional Segment Tree | | Chairman of the tree); Given an interval, M asks: the number of K-large numbers in the specified interval: the so-called chairman Tree, that is, the tree that can be persisted, that is, each time we insert a new element, we create a new node, so that all the historical versions of the segment tree can be preserved. Then consider the line tree subtraction, the two segment tree subtraction is each node subtraction, then we each node update once, then each element in the sequence corresponds to a version of the segment tree, that is, all the prefixes in the sequence of the weight of the segment tree, then for an interval, by the prefix subtraction will quickly Can get out of this section of the corresponding segment tree, and then ask this line tree of the K-value Note: If the interval K decimal value is converted to a large value (l-r-k+1) can be */#include <stdio.h> #include <string.h># Include <algorithm> #include <iostream>using namespace std;const int n=100010;int t[n];int num[n];int san[n]    ; int ls[n*20];int rs[n*20];int sum[n*20];int tot,rt;int n,m;void Build (int l,int r,int &rt) {rt=++tot;    sum[rt]=0;    if (l==r) return;    int m= (L+R)/2;    Build (L,m,ls[rt]); Build (M+1,r,rs[rt]);}    void update (int last,int p,int l,int r,int &rt) {rt=++tot;    Ls[rt]=ls[last];    Rs[rt]=rs[last];    sum[rt]=sum[last]+1;    if (l==r) return;    int m= (L+R) >>1;    if (p<=m) update (LS[LAST],P,L,M,LS[RT]); else update (RS[LAST],P,M+1,R,RS[RT]);} Int Query (int ss,int tt,int l,int r,int k) {if (l==r) return l;    int m= (L+R) >>1;    int CNT=SUM[LS[TT]]-SUM[LS[SS]];    if (k<=cnt) return query (LS[SS],LS[TT],L,M,K); else return query (RS[SS],RS[TT],M+1,R,K-CNT);}    int main () {int TT;    scanf ("%d", &AMP;TT);        while (tt--) {scanf ("%d%d", &n,&m);            for (int i=1; i<=n; i++) {scanf ("%d", &num[i]);        San[i]=num[i];        } tot=0;        Sort (san+1,san+n+1);        int Cnt=unique (san+1,san+n+1)-san-1;        Build (1,cnt,t[0]);        for (int i=1; i<=n; i++) {num[i]=lower_bound (san+1,san+1+cnt,num[i])-san;        } for (int i=1; i<=n; i++) update (t[i-1],num[i],1,cnt,t[i]);            while (m--) {int l,r,k;            scanf ("%d%d%d", &l,&r,&k);            int Id=query (T[L-1],T[R],1,CNT,K);        printf ("%d\n", San[id]); }} return 0;}

Hdu 2665 to persist segment tree for interval k large value (functional Segment Tree | | Chairman Tree)