POJ 2104 K-th Number [主席树入门]【数据结构】

题目链接:http://poj.org/problem?id=2104
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K-th Number
Time Limit: 20000MS Memory Limit: 65536K
Total Submissions: 53824 Accepted: 18506
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(2,3). The segment a[2…5] is (5,3). If we sort this segment,we get (2,6),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,and k (1 <= i <= j <= n,1 <= k <= j - i + 1) and represents the question Q(i,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.

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题目大意：
求区间第k大的数值

解题思路:
算法不用想 妥妥的主席树啊

之前一直没有学习主席树,就是理解不了主席树,
今天找到了一个很好的
博客1
博客2

算是明白了什么是主席树,

主席树为了保存历史版本的状态,于是在每次更新的时候都在开出一条链来代表新建出来的节点,

这样就不用每次在建一颗树了

其实主席树是一种函数式线段树

寻找$[l,r]$区间的值其实也就成了找 第$R$个版本-第$l-1$个版本 就是我们要查询的值了

附本题代码
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#pragma comment(linker,"/STACK:1024000000,1024000000")
//#include<bits/stdc++.h>
#include <stdio.h>
#include <algorithm>
using namespace std;

typedef long long int LL;

const int    INF = (~(1<<31));
const int    N   = 100000+7;
const double eps = 1e-7;

inline int read(){
    int x=0,f=1;char ch = getchar();
    while(ch<'0'||ch>'9'){if(ch=='-')f=-1;ch=getchar();}
    while('0'<=ch&&ch<='9'){x=(x<<3)+(x<<1)+ch-'0';ch=getchar();}
    return x*f;
}
/****************************************************/

int rt[N*20],ls[N*20],rs[N*20],sum[N*20],tot;
int n,q,ql,qr,x,a[N],b[N],sz;
void build(int &rt,int l,int r){
    rt=++tot;
    sum[rt]=0;
    if(l==r) return ;
    int m = (r+l)>>1;
    build(ls[rt],l,m);
    build(rs[rt],m+1,r);
}
void update(int &rt,int r,int last,int pos){
    rt = ++tot;
    ls[rt]=ls[last];
    rs[rt]=rs[last];
    sum[rt]=sum[last]+1;
    if(l==r) return ;
    int m = (r+l)>>1;
    if(pos<=m) update(ls[rt],m,ls[last],pos);
    else       update(rs[rt],r,rs[last],pos);
}
int query(int ss,int tt,int k){
    if(l==r)return l;
    int m = (l+r)>>1;
    int cnt=sum[ls[tt]]-sum[ls[ss]];
    if(k<=cnt)return query(ls[ss],ls[tt],k);
    else  return query(rs[ss],rs[tt],k-cnt);
}

void qq(){
    scanf("%d%d%d",&ql,&qr,&x);
    printf("%d\n",b[query(rt[ql-1],rt[qr],1,sz,x)]);
}

int main(){
    while(~scanf("%d%d",&n,&q)){

        for(int i=1;i<=n;i++) scanf("%d",a+i),b[i]=a[i];
        sort(b+1,b+n+1);
        sz=unique(b+1,b+n+1)-(b+1);

        tot = 0;
        build(rt[0],sz);

        for(int i=1;i<=n;i++)a[i]=lower_bound(b+1,b+sz+1,a[i])-b;
        for(int i=1;i<=n;i++)update(rt[i],rt[i-1],a[i]);
        while(q--)qq();
    }
    return 0;
}