poj 2104 主席树（区间第k大）

 

K-th Number

Time Limit: 20000MS		Memory Limit: 65536K
Total Submissions: 44940		Accepted: 14946
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

 

因为保存的是前i个数的情况，求[a,b]区间时，只需要T[b]-T[a-1]

/*
poj2104
主席树-求解区间第k大
hhh-2016-02-18 13:09:24
*/

#include <functional>
#include <iostream>
#include <cstdio>
#include <cstdlib>
#include <cstring>
#include <algorithm>
#include <cmath>
using namespace std;
typedef long long ll;
typedef long double ld;

using namespace std;
const int maxn = 100010;
int tot;
int n,m;
int a[maxn],t[maxn];
int T[maxn],lson[maxn*30],rson[maxn*30],c[maxn*30];

int build(int l,int r)
{
    int root = tot++;
    if(l != r)
    {
        int mid = (l+r)>>1;
        lson[root] = build(l,mid);
        rson[root] = build(mid+1,r);
    }
    return root;
}

void ini_hash()  //排序去重
{
    for(int i = 1;i <= n;i++)
        t[i] = a[i];
    sort(t+1,t+n+1);
    m = unique(t+1,t+n+1)-(t+1);
}

int Hash(int x) //得到位置
{
    return lower_bound(t+1,t+m+1,x)-t;
}

//如果那里发生改变则新建一个节点而非像平常修改那个节点的值
int update(int root,int pos,int val)
{
    int newroot = tot++;
    int tmp = newroot;
    c[newroot] = c[root] + val;
    int l = 1,r = m;
    while(l < r)
    {
        int mid = (l+r)>>1;
        if(pos <= mid)
        {
            lson[newroot] = tot++;rson[newroot] = rson[root];
            newroot = lson[newroot]; root = lson[root];
            r = mid;
        }
        else
        {
            lson[newroot] = lson[root]; rson[newroot] = tot++;
            newroot = rson[newroot]; root = rson[root];
            l = mid+1;
        }
        c[newroot] = c[root] + val;
    }
    return tmp;
}

int query(int lt,int rt,int k)
{
    int l = 1, r = m;
    while(l < r)
    {
         int mid = (l+r)>>1;
         if(c[lson[rt]] - c[lson[lt]] >= k)
         {
             r = mid;
             lt = lson[lt];
             rt = lson[rt];
         }
         else
         {
             l = mid+1;
             k -= (c[lson[rt]] - c[lson[lt]]);
             lt = rson[lt];
             rt = rson[rt];
         }
    }
    return l;
}

int main()
{
    int q;
    while(scanf("%d%d",&n,&q) !=EOF)
    {
        tot = 0;
        for(int i = 1;i <= n;i++)
            scanf("%d",&a[i]);
        ini_hash();
        T[0] = build(1,m);
        for(int i = 1;i <= n;i++)
        {
            int pos = Hash(a[i]);
            T[i] = update(T[i-1],pos,1);
        }
        while(q--)
        {
            int l,r,k;
            scanf("%d%d%d",&l,&r,&k);
            printf("%d\n",t[query(T[l-1],T[r],k)]);
        }
    }
    return 0;
}