POJ 2104 K-th Number 主席树(区间第k大)

题目链接：

http://poj.org/problem?id=2104

K-th Number

Time Limit: 20000MS
Memory Limit: 65536K #### 问题描述 > 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. #### 输入 > 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). #### 输出 > For each question output the answer to it --- the k-th number in sorted a[i...j] segment. ####样例输入 > 7 3 > 1 5 2 6 3 7 4 > 2 5 3 > 4 4 1 > 1 7 3 ####样例输出 > 5 > 6 > 3 ## 题意 > 给你n个数m个询问，每个询问给你l,r,k，求（l,r）区间里第k大的数，保证每个数只出现一次。

题解

主席树裸板。
一个简洁形象的教程：[port]
代码：[port]

代码

本土化：

#include<map>
#include<set>
#include<cmath>
#include<queue>
#include<stack>
#include<ctime>
#include<vector>
#include<cstdio>
#include<string>
#include<bitset>
#include<cstdlib>
#include<cstring>
#include<iostream>
#include<algorithm>
#include<functional>
#include<sstream>
using namespace std;
#define X first
#define Y second
#define mkp make_pair
#define lson(i) (tre[(i)].ls)
#define rson(i) (tre[(i)].rs)
#define sumv(i) tre[(i)].sum
#define mid (l+(r-l)/2)
#define sz() size()
#define pb(v) push_back(v)
#define all(o) (o).begin(),(o).end()
#define clr(a,v) memset(a,v,sizeof(a))
#define bug(a) cout<<#a<<" = "<<a<<endl
#define rep(i,a,b) for(int i=a;i<(b);i++)
#define scf scanf
#define prf printf

typedef long long LL;
typedef vector<int> VI;
typedef pair<int,int> PII;
typedef vector<pair<int,int> > VPII;

const int INF=0x3f3f3f3f;
const LL INFL=0x3f3f3f3f3f3f3f3fLL;
const double eps=1e-8;
const double PI = acos(-1.0);

//start----------------------------------------------------------------------

const int maxn=1e5+10;

///nlogn空间复杂度
struct Tre{
    int ls,rs,sum;
    Tre(){ls=rs=sum=0;}
}tre[maxn*20];

int n,m;
int rt[maxn],tot;

int _v;
void update(int &o,int l,int r){
    tre[++tot]=tre[o],o=tot;
    if(l==r){
        sumv(o)++;
    }else{
        if(_v<=mid) update(lson(o),l,mid);
        else update(rson(o),mid+1,r);
        sumv(o)=sumv(lson(o))+sumv(rson(o));
    }
}

int _res;
void query(int o1,int o2,int l,int r,int k){
    if(l==r){
        _res=l;
    }else{
        ///前缀和思想
        int cnt=sumv(lson(o2))-sumv(lson(o1));
        if(cnt>=k) query(lson(o1),lson(o2),l,mid,k);
        else query(rson(o1),rson(o2),mid+1,r,k-cnt);
    }
}

int idx[maxn],arr[maxn],ra[maxn];

bool cmp(int x,int y){
    return arr[x]<arr[y];
}

///0是个超级节点
void init(){
    rt[0]=tot=0;
}

int main() {
    while(scf("%d%d",&n,&m)==2&&n){
        init();
        for(int i=1;i<=n;i++) scf("%d",&arr[i]);
        for(int i=1;i<=n;i++) idx[i]=i;

        ///离散化
        sort(idx+1,idx+n+1,cmp);
        for(int i=1;i<=n;i++) ra[idx[i]]=i;

        ///主席树
        for(int i=1;i<=n;i++){
            _v=ra[i];
            rt[i]=rt[i-1];
            update(rt[i],1,n);
        }

        ///查询区间第k大
        while(m--){
            int l,r,k;
            scf("%d%d%d",&l,&r,&k);
            query(rt[l-1],rt[r],1,n,k);
            prf("%d\n",arr[idx[_res]]);
        }
    }
    return 0;
}

//end-----------------------------------------------------------------------