POJ-2104 主席树板子

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
Hint
This problem has huge input,so please use c-style input(scanf,printf),or you may got time limit exceed.

每加入一个点i建一棵线段树root[i]
那么线段树 root[r]-root[l-1] 的控制范围是[l,r]
查询第k大，如果l==r,return l
如果k大于左子树权值s，则查询右子树k-s
否则继续查询左子树
板子

#include
#include
#include
#include
#include
#include
using namespace std;
const int N=1e5+7;
struct node{int l,r,sum;}T[N<<5];
int n,m,cnt;
int root[N],a[N];
vector<int>v;
int id(int x){return lower_bound(v.begin(),v.end(),x)-v.begin()+1;}
void update(int &x,int y,int pos,int l=1,int r=n){
	T[++cnt]=T[y],++T[x=cnt].sum;
	if(l==r)return;
	int mid=(l+r)>>1;
	if(mid>=pos)update(T[x].l,T[y].l,pos,l,mid);
	else update(T[x].r,T[y].r,pos,mid+1,r);
}
int query(int x,int y,int k,int l=1,int r=n){
	if(l==r)return l;
	int mid=(l+r)>>1;
	int sum=T[T[y].l].sum-T[T[x].l].sum;
	if(sum>=k)return query(T[x].l,T[y].l,k,l,mid);
	else return query(T[x].r,T[y].r,k-sum,mid+1,r);
}
int main(){
	scanf("%d%d",&n,&m);
	for(int i=1;i<=n;++i)scanf("%d",a+i),v.push_back(a[i]);
	sort(v.begin(),v.end());
	v.resize(unique(v.begin(),v.end())-v.begin());
	for(int i=1;i<=n;++i)update(root[i],root[i-1],id(a[i]));
	while(m--){
		int l,r,k;
		scanf("%d%d%d",&l,&r,&k);
		printf("%d\n",v[query(root[l-1],root[r],k)-1]);
	}
	return 0;
}