poj 2388 Heap Sort Heap Median

Who's in the Middle

Time Limit: 1000MS Memory Limit: 65536KTotal Submissions: 29341 Accepted: 17008

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

524135

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.

A standard method to build a heap:

//1. 利用宏定义swap//2. max_heapify(largest,heap,len);开始忘了递归了，自底向上地递归 #include<stdio.h>#include<iostream>#include<stdlib.h>#include<algorithm>using namespace std;int N,milk[1000001],size;void max_heapify(int i,int heap[],int len){//数组从0开始   int l,r,largest;  l=2*i+1;   r=2*i+2;  if( l<len && heap[l]>heap[i] )    largest=l;  else    largest=i;  if( r<len && heap[r]>heap[largest] )    largest=r;  if(largest!=i){    swap( heap[largest],heap[i] );    max_heapify(largest,heap,len);  }}void parent_heapify(int i,int heap[],int len){//和parent结点不断交换   int parent = i / 2;  while (heap[i] > heap[parent]) {    swap(heap[i], heap[parent]);    i = parent, parent = i / 2;  }}int main(){  int i,j;  scanf("%d",&size);  for(i=0;i<=size/2;i++){    scanf("%d",&milk[i]);  }  for(i=(size/2+1)/2-1;i>=0;i--)    max_heapify(i,milk,size/2+1);  for(i=size/2+1;i<size;i++){    scanf("%d",&milk[i]);    if(milk[i]<milk[0]){      swap( milk[i],milk[0] );      max_heapify(0,milk,size/2+1);                 }  }  printf("%d\n",milk[0]);  //system("pause");  return 0; }