poj 2104 K-th Number 主席树

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.
Source
Northeastern Europe 2004, Northern Subregion

大意：看样例懂题意，显然给你n(n<=1e5)个数，每个数小于1e9,给你m(m<=5000)次询问，求一个区间第k大。

分析：模板题。把1-i建可持久化线段数，把两棵相减跑出解，要先对a进行离散。（也就是主席树水题）

代码：

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

const int maxn= 1e5+7;

using namespace std;


struct node{
    int l,r,sum;
};

node t[maxn*40];
int root[maxn],cnt,n,m,a[maxn],num[maxn],rank[maxn],x,y,k;
int i;

void kp(int l,int r)
{
    if (l>r) return;
    int i=l; int j=r;
    int temp;
    int key=a[(l+r)/2];
    while (i<=j)
    {
        while (a[i]<key) i++;
        while (a[j]>key) j--;
        if (i<=j)
        {
            temp=a[i]; a[i]=a[j]; a[j]=temp;
            temp=num[i]; num[i]=num[j]; num[j]=temp;
            i++; j--;
        }
    }
    kp(l,j);
    kp(i,r);
}

void updata(int l,int r,int &x,int y,int c)
{
    t[++cnt]=t[y];t[cnt].sum++;x=cnt;
    if (l==r) return;
    int mid=(l+r)/2;
    if (mid>=c) updata(l,mid,t[x].l,t[y].l,c);
    else updata(mid+1,r,t[x].r,t[y].r,c);
} 

int ask(int l,int r,int x,int y,int c)
{
    if (l==r)
    {
        return l;
    }
    int mid=(l+r)/2; 
    int sum=t[t[y].l].sum-t[t[x].l].sum;
    if (sum>=c) return ask(l,mid,t[x].l,t[y].l,c);
           else return ask(mid+1,r,t[x].r,t[y].r,c-sum);
}

int main()
{
    scanf("%d%d",&n,&m);
    for (i=1;i<=n;i++)
    {
        scanf("%d",&a[i]);
        num[i]=i;
    }   
    kp(1,n);
    for (i=1;i<=n;i++)
        rank[num[i]]=i;     
    for (i=1;i<=n;i++)
    {
        updata(1,n,root[i],root[i-1],rank[i]);
    }   
    for (i=1;i<=m;i++)
    {
        scanf("%d%d%d",&x,&y,&k);
        int ans=ask(1,n,root[x-1],root[y],k);
        printf("%d\n",a[ans]);
    }
}