POJ 2388 Who's in the Middle(堆排序)

Who's in the Middle

Time Limit: 1000MS		Memory Limit: 65536K
Total Submissions: 22341		Accepted: 12765

Description

FJ is surveying his herd to find the most average cow. He wants to know how much milk this 'median' cow gives: half of the cows give as much or more than the median; half give as much or less. 

Given an odd number of cows N (1 <= N < 10,000) and their milk output (1..1,000,000), find the median amount of milk given such that at least half the cows give the same amount of milk or more and at least half give the same or less.

Input

* Line 1: A single integer N 

* Lines 2..N+1: Each line contains a single integer that is the milk output of one cow.

Output

* Line 1: A single integer that is the median milk output.

Sample Input

5

2

4

1

3

5

Sample Output

3

Hint

INPUT DETAILS: 

Five cows with milk outputs of 1..5 

OUTPUT DETAILS: 

1 and 2 are below 3; 4 and 5 are above 3.

Source

USACO 2004 November

 解题报告：这就是最简单的排序问题，求中位数，为了练习堆排序，就建立了小根堆实现的，

代码如下：

#include <iostream>
#include <cstdio>
#include <cstring>
#include <cstdlib>
using namespace std;
const int N = 10010;
int a[N], b[N];
void Head_adjust(int x, int y)//建立小根堆
{
    int i, j, temp;
    i = x;
    j = 2 * i;
    temp = a[i];
    while (j <= y)
    {
        //if (j < y && a[j + 1] > a[j])//这是建立大根堆
        if (j < y && a[j + 1] < a[j])
        {
            ++ j;
        }
        //if (temp < a[j])//这是建立大根堆
        if (temp > a[j])
        {
            a[i] = a[j];
            i = j;
            j = 2 * i;
        }
        else
        {
            break;
        }
        a[i] = temp;
    }
}
int main()
{
    int n, i, mid, m;
    scanf("%d", &n);
    for (i = 1; i <= n; ++i)
    {
        scanf("%d", &a[i]);
    }
    for (i = n/2; i > 0;--i)
    {
        Head_adjust(i, n);//将二叉树变成小根堆
    }
    m = n;
    i = 0;
    while (n --)//每次都取堆得第一个元素
    {
        b[i] = a[1];
        i ++;
        a[1] = a[n + 1];
        Head_adjust(1, n);//调整
    }
    mid = m / 2;//因为b[]是从b[0]开始存储的，所以直接除以二就行
    printf("%d\n", b[mid]);
    return 0;
}