POJ 2104 K-th Number——————主席树模板，区间第k大

K-th Number

Time Limit: 20000MS Memory Limit: 65536K
Total Submissions: 77307 Accepted: 27874
Case Time Limit: 2000MS

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.

模板题，求区间第k大

#include<cstdio>
#include<vector>
#include<iostream>
#include<algorithm>

using namespace std;
/*
求区间第k大
poj - 2104
*/

const int MAXN = 1e5+7;
int n, m, cnt, root[MAXN], a[MAXN], x, y, k;
struct node{int l, r, sum;} T[MAXN*40];
vector<int> v;
int getid(int x) { return lower_bound(v.begin(), v.end(), x)-v.begin() + 1;}

void update(int l, int r, int &x, int y, int pos) {// T[y] 是上一版本的， T[x] 是这一版本的
        T[++cnt] = T[y]; T[cnt].sum++; x = cnt;
        if(l == r) return ;
        int mid = (l+r)>>1;
        if(mid >= pos)  update(l, mid, T[x].l, T[y].l, pos);
        else            update(mid+1, r, T[x].r, T[y].r, pos);
}
int query(int l, int r, int x, int y, int k) {
        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(l, mid, T[x].l, T[y].l, k);
        else            return query(mid+1, r, T[x].r, T[y].r, k - sum);
}

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.erase(unique(v.begin(), v.end()), v.end());
        for(int i=1;i<=n;++i)   update(1,n,root[i],root[i-1],getid(a[i]));
        for(int i=1;i<=m;++i) {
                scanf("%d %d %d",&x,&y,&k);
                printf("%d\n",v[query(1, n, root[x-1], root[y], k) - 1]);
        }
        return 0;
}