POJ_3519 Median

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本文介绍了一种通过二分查找结合排序的方法，快速找出给定数字集合中所有两两差值的中位数。该方法适用于大规模数据集，通过不断调整候选中位数直至找到符合条件的值。

Median

题目链接

POJ_3519 Median

Description

Given $N$ numbers, $X 1, X 2, . . ., X N$ , let us calculate the difference of every pair of numbers: $∣ X i - X j ∣ (1 \leq i ＜ j \leq N)$ . We can get $C (N, 2)$ differences through this work, and now your task is to find the median of the differences as quickly as you can!

Note in this problem, the median is defined as the $(m / 2) - t h$ smallest number if m,the amount of the differences, is even. For example, you have to find the third smallest one in the case of m = 6.

Input

The input consists of several test cases.
In each test case, $N$ will be given in the first line. Then N numbers are given, representing $X_1, X_2, ... , X_N, ( X_i ≤ 1,000,000,000 3 ≤ N ≤ 1,00,000 )$

Output

For each test case, output the median in a separate line.

Sample Input

Sample Output

1
8

解析

本题中，数列中数的顺序对结果不会有影响。

我们是使用二分答案来求中位数的值。先对数列进行排序。接着，分别枚举 $∣ X i - X j ∣ (1 \leq i ＜ j \leq N)$ 中的 $i$ 与 $j$ ，求出有多少个 $∣ X i - X j ∣$ 的值大于等于（或小于）二分值，再根据这个结果继续二分。

代码

#include<algorithm>
#include<cstdio>
#include<queue>
#include<iostream>
#include<cmath>
using namespace std;
long long n,a[100001],l,r,mid,cnt,maxx,ans;
int main()
{
	while(scanf("%lld",&n)!=EOF)
	{
		maxx=n*(n-1)/2;
		for(int i=1;i<=n;i++)
			scanf("%lld",&a[i]);
		sort(a+1,a+n+1);
		l=0,r=2147483647;
		while(l<=r)//二分答案求中位数 
		{
			cnt=0;
			mid=(l+r)/2;
			int j=1;//左端点 
			for(int i=2;i<=n;i++)//右端点 
			{
				while(a[i]-a[j]>=mid)
					j++;
				cnt+=i-j;
			}
			if(cnt<maxx/2+maxx%2)//中位数太小 
			{
				ans=mid;
				l=mid+1;
			} 
			else//中位数太大 
				r=mid-1;
		}
		printf("%lld\n",ans);
	}
	return 0;
}