POJ 2104 K-th Number

POJ 2104 K-th Number

题目链接

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

主席树模板题，查询区间第 $k$ 大的值，AC代码如下:

#include
#include
#include
using namespace std;
const int N=2e5+5;

int n,m,siz,l,r,k,cnt=0;
int a[N];//原数组
int s[N];//去重后的数组
int rk[N];//记录原数组的元素的排名
int root[N];//记录每个根节点的编号

struct ptree{
     
    int l,r,sum;//sum记录经过该节点的次数
}t[N*20];

void build(int &node,int l,int r){
     
    node=++cnt;
    if(l==r) return;
    int mid=l+r>>1;
    build(t[node].l,l,mid);
    build(t[node].r,mid+1,r);
}

void update(int &node,int last,int l,int r,int s){
     
    node=++cnt;
    t[node]=t[last];
    ++t[node].sum;
    if(l==r) return;
    int mid=l+r>>1;
    if(s<=mid) update(t[node].l,t[last].l,l,mid,s);
    else update(t[node].r,t[last].r,mid+1,r,s);
}

int query(int node,int last,int l,int r,int k) {
     
    if (l == r) return s[l];
    int sum = t[t[node].l].sum - t[t[last].l].sum, mid = l + r >> 1;
    if (k <= sum) return query(t[node].l, t[last].l, l, mid, k);
    else return query(t[node].r, t[last].r, mid + 1, r, k - sum);
}

int main(){
     
    scanf("%d%d",&n,&m);
    for(int i=1;i<=n;i++) scanf("%d",&a[i]);
    memcpy(s,a,sizeof(s));
    sort(s+1,s+1+n);
    siz=unique(s+1,s+1+n)-s-1;
    build(root[0],1,siz);
    for(int i=1;i<=n;i++) rk[i]=lower_bound(s+1,s+1+siz,a[i])-s;
    for(int i=1;i<=n;i++) update(root[i],root[i-1],1,siz,rk[i]);
    while(m--){
     
        scanf("%d%d%d",&l,&r,&k);
        printf("%d\n",query(root[r],root[l-1],1,siz,k));
    }
}