POJ 2761 Feed the dogs（平衡树or划分树or主席树）

Description

Wind loves pretty dogs very much, and she has n pet dogs. So Jiajia has to feed the dogs every day for Wind. Jiajia loves Wind, but not the dogs, so Jiajia use a special way to feed the dogs. At lunchtime, the dogs will stand on one line, numbered from 1 to n, the leftmost one is 1, the second one is 2, and so on. In each feeding, Jiajia choose an inteval[i,j], select the k-th pretty dog to feed. Of course Jiajia has his own way of deciding the pretty value of each dog. It should be noted that Jiajia do not want to feed any position too much, because it may cause some death of dogs. If so, Wind will be angry and the aftereffect will be serious. Hence any feeding inteval will not contain another completely, though the intervals may intersect with each other. 
Your task is to help Jiajia calculate which dog ate the food after each feeding. 

Input

The first line contains n and m, indicates the number of dogs and the number of feedings. 
The second line contains n integers, describe the pretty value of each dog from left to right. You should notice that the dog with lower pretty value is prettier. 
Each of following m lines contain three integer i,j,k, it means that Jiajia feed the k-th pretty dog in this feeding. 
You can assume that n<100001 and m<50001. 

Output

Output file has m lines. The i-th line should contain the pretty value of the dog who got the food in the i-th feeding.

 

题目大意：给n个数，m次询问区间[l, r]的第k小数。

思路：对询问排序，再离线处理每一个询问（加减点直至平衡树对应区间等于询问区间），不过这样做复杂度是没有保障的……因此差不多的题目如POJ2104妥妥地TLE了……

关于正解之划分树和主席树这里不写。

关于划分树和主席树：

POJ 2104 K-th Number（划分树）

POJ 2104 K-th Number（不带修改主席树——附讲解）

 

代码（3438MS）（treap树）：

#include <cstdio>
#include <cstring>
#include <iostream>
#include <algorithm>
using namespace std;

const int MAXN = 100010;

int key[MAXN], weight[MAXN], child[MAXN][2], size[MAXN];
int stk[MAXN], top, cnt;

inline void init() {
    top = cnt = 0;
}

inline int new_node(int k) {
    int x = (top ? stk[top--] : ++cnt);
    key[x] = k;
    size[x] = 1;
    weight[x] = rand();
    child[x][0] = child[x][1] = 0;
    return x;
}

inline void update(int &x) {
    size[x] = size[child[x][0]] + size[child[x][1]] + 1;
}

void rotate(int &x, int t) {
    int y = child[x][t];
    child[x][t] = child[y][t ^ 1];
    child[y][t ^ 1] = x;
    update(x); update(y);
    x = y;
}

void insert(int &x, int k) {
    if(x == 0) x = new_node(k);
    else {
        int t = key[x] < k;
        insert(child[x][t], k);
        if(weight[child[x][t]] < weight[x]) rotate(x, t);
    }
    update(x);
}

void remove(int &x, int k) {
    if(key[x] == k) {
        if(child[x][0] && child[x][1]) {
            int t = weight[child[x][0]] < weight[child[x][1]];
            rotate(x, t); remove(child[x][t ^ 1], k);
        }
        else {
            stk[++top] = x;
            x = child[x][0] + child[x][1];
        }
    }
    else remove(child[x][key[x] < k], k);
    if(x > 0) update(x);
}

int kth(int &x, int k) {
    if(x == 0 || k <= 0 || k > size[x]) return 0;
    int s = 0;
    if(child[x][0]) s = size[child[x][0]];
    if(k == s + 1) return key[x];
    if(k <= s) return kth(child[x][0], k);
    return kth(child[x][1], k - s - 1);
}

struct Node {
    int l, r, k, id;
    void read(int i) {
        id = i;
        scanf("%d%d%d", &l, &r, &k);
    }
    bool operator < (const Node &rhs) const {
        if(r != rhs.r) return r < rhs.r;
        return l < rhs.l;
    }
};

Node query[MAXN];
int a[MAXN], ans[MAXN];

int main() {
    int n, m;
    scanf("%d%d", &n, &m);
    for(int i = 1; i <= n; ++i) scanf("%d", &a[i]);
    for(int i = 0; i < m; ++i) query[i].read(i);
    sort(query, query + m);
    int left = 1, right = 0, root = 0;
    for(int i = 0; i < m; ++i) {
        while(right < query[i].r)
            if(!root) root = new_node(a[++right]);
            else insert(root, a[++right]);
        while(left > query[i].l)
            if(!root) root = new_node(a[--left]);
            else insert(root, a[--left]);
        while(left < query[i].l) remove(root, a[left++]);
        ans[query[i].id] = kth(root, query[i].k);
    }
    for(int i = 0; i < m; ++i) printf("%d
", ans[i]);
}