题目链接：

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

K-th Number

Time Limit: 20000MSMemory 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 printftypedef 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-----------------------------------------------------------------------