【主席树】POJ 2014 K-th Number

POJ 2014 K-th Number

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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.

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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).

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Output

For each question output the answer to it — the k-th number in sorted a[i…j] segment.

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Sample Input

7 3
1 5 2 6 3 7 4
2 5 3
4 4 1
1 7 3

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Sample Output

5
6
3

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HINT

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

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Solution

题意就是给你一个数组然后求其中一段区间的第k大
离散一下吧。。
然后貌似可以分块？(并不行)
这里介绍另外一种算法

主席树:
一种神奇的方法
首先每个节点建一棵树,每个点权值是前k个元素在该点管辖范围[L,R]之间的元素个数(也可以说前缀和)
然而这样建树是n^2logn的空间?
我们发现,每次加一个元素,只会修改树的左子树或者右子树.
所以每次使用一下之前的树即可,空间优化到nlogn.
询问[l,r]区间第k大时二分一下即可.

Code

#include
#include
#include
#include

using namespace std;

#define maxn 100001

struct node{
    int l,r,v;
}t[maxn*20];

inline int in()
{
    int x=0,f=1;char ch=getchar();
    while((ch<'0'||ch>'9')&&ch!='-')ch=getchar();
    if(ch=='-')ch=getchar(),f=-1;
    while(ch<='9'&&ch>='0')x=x*10+ch-'0',ch=getchar();
    return x*f;
}

int a[maxn],p[maxn],b[maxn],size,root[maxn];

bool cmp(const int A,const int B){return a[A]void update(int &poi,int l,int r,int num)
{
    t[++size]=t[poi];poi=size;
    t[size].v++;
    if(l==r)return;
    int mid=(l+r)>>1;
    if(num>mid)update(t[poi].r,mid+1,r,num);
    else update(t[poi].l,l,mid,num);
}

int query(int L,int R,int l,int r,int k)
{
    if(l==r)return l;
    int T=t[t[R].l].v-t[t[L].l].v;
    int mid=(l+r)>>1;
    if(T>=k)return query(t[L].l,t[R].l,l,mid,k);
    else return query(t[L].r,t[R].r,mid+1,r,k-T);
}

int main()
{
    freopen("2104.in","r",stdin);
    int n,m,x,y,k;
    n=in(),m=in();
    for(int i=1;i<=n;i++)a[i]=in(),p[i]=i;
    sort(1+p,p+n+1,cmp);
    for(int i=1;i<=n;i++)b[p[i]]=i;
    for(int i=1;i<=n;i++)
    {
        root[i]=root[i-1];
        update(root[i],1,n,b[i]);
    }
    for(int i=1;i<=m;i++)
    {
        x=in(),y=in(),k=in();
        int X=query(root[x-1],root[y],1,n,k);
        printf("%d\n",a[p[X]]);
    }
    return 0;
}