POJ 2166 Heapsort（递推）

Description

A well known algorithm called heapsort is a deterministic sorting algorithm taking O(n log n) time and O(1) additional memory. Let us describe ascending sorting of an array of different integer numbers.  The algorithm consists of two phases. In the first phase, called heapification, the array of integers to be sorted is converted to a heap. An array a[1 . . . n] of integers is called a heap if for all 1 <= i <= n the following heap conditions are satisfied: 

• if 2i <= n then a[i] > a[2i]; 
• if 2i + 1 <= n then a[i] > a[2i + 1].

We can interpret an array as a binary tree, considering children of element a[i] to be a[2i] and a[2i + 1]. In this case the parent of a[i] is a[i div 2], where i div 2 = [i/2]. In terms of trees the property of being a heap means that for each node its value is greater than the values of its children.  In the second phase the heap is turned into a sorted array. Because of the heap condition the greatest element in the heapified array is a[1]. Let us exchange it with a[n], now the greatest element of the array is at its correct position in the sorted array. This is called extract-max.  Now let us consider the part of the array a[1 . . . n-1]. It may be not a heap because the heap condition may fail for i = 1. If it is so (that is, either a[2] or a[3], or both are greater than a[1]) let us exchange the greatest child of a[1] with it, restoring the heap condition for i = 1. Now it is possible that the heap condition fails for the position that now contains the former value of a[1]. Apply the same procedure to it, exchanging it with its greatest child. Proceeding so we convert the whole array a[1 . . . n-1] to a heap. This procedure is called sifting down. After converting the part a[1 . . . n-1] to a heap by sifting, we apply extract-max again, putting second greatest element of the array to a[n - 1], and so on.  For example, let us see how the heap a = (5, 4, 2, 1, 3) is converted to a sorted array. Let us make the first extract-max. After that the array turns to (3, 4, 2, 1, 5). Heap condition fails for a[1] = 3 because its child a[2] = 4 is greater than it. Let us sift it down, exchanging a[1] and a[2]. Now the array is (4, 3, 2, 1, 5). The heap condition is satisfied for all elements, so sifting is over. Let us make extract-max again. Now the array turns to (1, 3, 2, 4, 5). Again the heap condition fails for a[1]; exchanging it with its greatest child we get the array (3, 1, 2, 4, 5) which is the correct heap. So we make extract-max and get (2, 1, 3, 4, 5). This time the heap condition is satisfied for all elements, so we make extract-max, getting (1, 2, 3, 4, 5). The leading part of the array is a heap, and the last extract-max finally gives (1, 2, 3, 4, 5).  It is known that heapification can be done in O(n) time. Therefore, the most time consuming operation in heapsort algorithm is sifting, which takes O(n log n) time. In this problem you have to find a heapified array containing different numbers from 1 to n, such that when converting it to a sorted array, the total number of exchanges in all sifting operations is maximal possible. In the example above the number of exchanges is 1 + 1 + 0 + 0 + 0 = 2, which is not the maximum. (5, 4, 3, 2, 1) gives the maximal number of 4 exchanges for n = 5.

Input

Input contains n (1 <= n <= 50 000).

Output

Output the array containing n different integer numbers from 1 to n, such that it is a heap, and when converting it to a sorted array, the total number of exchanges in sifting operations is maximal possible. Separate numbers by spaces.

 

题目大意：构造一个大根堆，使得用这个大根堆从小到大排序的时候交换次数最多。

思路：对于一个n个元素的大根堆，交换1和n，然后进行调整（即交换）。为了交换次数最多，我们希望它能交换到最底层，那个数当然最好是1。然后为了保证深度最大（有可能左边的深度比左边大1），那么我们希望顶元素1调整到n-1的位置。于是开始递推，对于 i-1 个元素最大的情况，最后一位是1，那么我们把1调整到最顶端。然后顶端置 i ，第 i 位置1。那么，我们进行排序到剩下 i 个元素的时候，顶端和位置n交换，剩下 i-1个元素，就必然需要把顶端的1调整到 i-1 的位置。因为我们把1往上调整的时候，比如要把x = heap[j/2]移动到heap[j]，那么当1调整到j/2的时候，x肯定是j/2比较大的儿子（因为另一个儿子曾经是x的儿子）。

PS：我用G++交TLE了从C++交AC了你们看着办吧……

 

代码（391MS）：

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

const int MAXN = 50010;

int heap[MAXN];

int main() {
    int n;
    while(scanf("%d", &n) != EOF) {
        heap[1] = 1;
        for(int i = 2; i <= n; ++i) {
            for(int j = i - 1; j != 1; j >>= 1) {
                heap[j] = heap[j >> 1];
            }
            heap[1] = i;
            heap[i] = 1;
        }
        for(int i = 1; i < n; ++i) printf("%d ", heap[i]);
        printf("%d\n", heap[n]);
    }
}