信息竞赛学习笔记：POJ3579中位数（二分）

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本文介绍了一种高效算法，用于从给定的一组整数中计算所有可能数对差值的中位数。通过使用二分查找和快速排序技巧，该算法能在大数据集上快速运行。

Median

Time Limit: 1000MS		Memory Limit: 65536K
Total Submissions: 5066		Accepted: 1618

Description

Given N numbers, X₁, X₂, ... , X_N, let us calculate the difference of every pair of numbers: ∣X_i - X_j∣ (1 ≤ i ＜ j ≤ 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)-th 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 X₁, X₂, ... , 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

Source

POJ Founder Monthly Contest – 2008.04.13, Lei Tao

题的大概意思是给你N个数，两两做差取绝对值，再找出这些绝对值的中位数。老师的思路一开始是在求完差的快排中再次二分，然而我并没有学会，且这个N<100000，光求差就炸了，何况一个测试点还多组数据………老师之后提示我双重二分，在参考了一些思路后，我写出了一种二分。代码如下：

#include<iostream>
#include<string>
#include<algorithm>
#include<cstdio>
using namespace std;
const int maxn=110000;
int n,a[maxn],k,lp,rp,mid;
bool check(int mid)
{
	int sum=0,temp;
	for(int i=0;i<n;i++)
		{
			temp=a[i]-(mid-1);
			sum+=&a[i]-lower_bound(a,a+n,temp);
		}
	if(sum>=k) return false;
	else return true;
}
int main()
{
	while(scanf("%d",&n)!=EOF)
	{
		for(int i=0;i<n;i++)
		   scanf("%d",&a[i]);
		k=n*(n-1)/2;
		if(k%2==0) k=k/2;
		else k=k/2+1;
		sort(a,a+n);
		lp=0;rp=a[n-1]-a[0];
		while(rp-lp>1) 
		{
			mid=(lp+rp)/2;
			if(check(mid)) lp=mid;
			else rp=mid;
		}
		cout<<lp<<endl;
	}
	return 0;
}

这道题有几个坑点：

第一个：读数不能用cin，否则必超无疑！（我的AC啊

）