Online Judge 2014 K-th Number -主席树

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.InputThe 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 10 ⁹ 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).OutputFor 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.

---

　　很裸的主席树模板题。不知道可不可以用O（n^1.5log₂n）的莫队加平衡树来做。

　　如果用下面这段代码去过poj的话会T掉，如果不是用new动态分配内存而是直接分配一个数组，在数组中储存所有节点可能就可以过了。

Code

/**
 * OpenJudge
 * Bailian#2104
 * Accepted
 * Time:2247ms
 * Memory:65536k
 */
#include<iostream>
#include<fstream>
#include<sstream>
#include<cstdio>
#include<cstdlib>
#include<cstring>
#include<ctime>
#include<cctype>
#include<cmath>
#include<algorithm>
#include<stack>
#include<queue>
#include<set>
#include<map>
#include<vector>
using namespace std;
typedef bool boolean;
#define smin(a, b)    (a) = min((a), (b))
#define smax(a, b)    (a) = max((a), (b))
template<typename T>
inline void readInteger(T& u){
    char x;
    int aFlag = 1;
    while(!isdigit((x = getchar())) && x != '-' && x != -1);
    if(x == -1){
        ungetc(x, stdin); 
        return;
    }
    if(x == '-'){
        x = getchar();
        aFlag = -1;
    }
    for(u = x - '0'; isdigit((x = getchar())); u = (u << 3) + (u << 1) + x - '0');
    ungetc(x, stdin);
    u *= aFlag;
}

typedef class SegTreeNode{
    public:
        int s;    //size
        SegTreeNode *left, *right;
        SegTreeNode():left(NULL), right(NULL), s(0){        }
        void pushUp(){
            this->s += this->left->s + this->right->s;
        }
}SegTreeNode;

typedef class SegTree{
    public:
        SegTreeNode* root;
        SegTree():root(NULL){        }
        SegTree(int size, int val){
            build(root, 1, size, val);
        }
        SegTree(SegTreeNode* root):root(root){    }
        
        void build(SegTreeNode*& node, int from, int end, int val){
            node = new SegTreeNode();
            if(from == end){
                node->s = (val == from) ? (1) : (0);
                return;
            }
            int mid = (from + end) >> 1;
            build(node->left, from, mid, val);
            build(node->right, mid + 1, end, val);
            node->pushUp();
        }
}SegTree;

typedef class Pair{
    public:
        int data;
        int index;
        Pair(const int data = 0, const int index = 0):data(data), index(index){        }
        boolean operator < (Pair another) const {
            return data < another.data;
        }
}Pair;

typedef class ChairTree{    //主席树
    public: 
        SegTree* st;
        int* discrete;
        map<int, int> to;
        int len;
        ChairTree():st(NULL), discrete(NULL){        }
        ChairTree(int* lis, int len):len(len){
            discrete = new int[(const int)(len + 1)];
            memcpy(discrete, lis, sizeof(int) * (len + 1));
            sort(discrete + 1, discrete + len + 1);
            for(int i = 1; i <= len; i++)
                to[discrete[i]] = i;
            st = new SegTree[(const int)(len + 1)];
            st[0] = SegTree(len, 0);
            for(int i = 1; i <= len; i++){
                st[i] = SegTree();
                appendSegTree(st[i].root, st[i - 1].root, 1, len, to[lis[i]]);
            }
        }
        
        void appendSegTree(SegTreeNode*& node, SegTreeNode*& before, int from, int end, int val){
            node = new SegTreeNode();
            if(from == end){
                node->s = 1;
                return;
            }
            int mid = (from + end) >> 1;
            if(val <= mid){
                node->right = before->right;
                appendSegTree(node->left, before->left, from, mid, val);
            }else{
                node->left = before->left;
                appendSegTree(node->right, before->right, mid + 1, end, val);
            }
            node->pushUp();
        }
        
        int query(int l, int r, SegTreeNode*& left, SegTreeNode*& right, int k){
            if(l == r){
                return l;
            }
            int ls = left->left->s - right->left->s;
            int mid = (l + r) >> 1;
            if(ls >= k)    return query(l, mid, left->left, right->left, k);
            return query(mid + 1, r, left->right, right->right, k - ls);
        }
        
        int query(int from, int end, int k){
            return discrete[query(1, len, st[end].root, st[from - 1].root, k)];
        }
}ChairTree;

int n, m;
ChairTree ct;
int *lis;

inline void init(){
    readInteger(n);
    readInteger(m);
    lis = new int[(const int)(n + 1)];
    for(int i = 1; i <= n; i++){
        readInteger(lis[i]);
    }
    ct = ChairTree(lis, n);
}

inline void solve(){
    int l, r, k;
    while(m--){
        readInteger(l);
        readInteger(r);
        readInteger(k);
        int res = ct.query(l, r, k);
        printf("%d\n", res);
    }
}

int main(){
    init();
    solve();
    return 0;
}