信息競賽學習筆記：POJ3579中位數（二分）

原創

2020-02-21 11:07

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

題的大概意思是給你N個數，兩兩做差取絕對值，再找出這些絕對值的中位數。老師的思路一開始是在求完差的快排中再次二分，然而我並沒有學會，且這個N<100000，光求差就炸了，何況一個測試點還多組數據………老師之後提示我雙重二分，在參考了一些思路後，我寫出了一種二分。代碼如下：

#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;
}

這道題有幾個坑點：

第一個：讀數不能用cin，否則必超無疑！（我的AC啊

）