POJ 3579 median

最新推荐文章于 2019-02-11 19:08:16 发布

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最新推荐文章于 2019-02-11 19:08:16 发布

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分类专栏：二分

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本文介绍了一种高效算法，用于求解给定序列中任意两数之差构成的新序列的中位数。通过排序和二分查找技巧，避免了直接计算所有差异值的方法，显著提高了计算效率。

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Median

Time Limit: 1000MS		Memory Limit: 65536K
Total Submissions: 7493		Accepted: 2555

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

题目大意：有一个序列，将其任意两个数的差生成一个新的序列，求这个序列的中位数。

思路：数据范围太大，肯定不能一个一个算，所以二分答案,问题就变成了判定生成的这个序列中小于等于答案
的数的个数是否小于等于n*(n-1)/4.具体做法：首先将原数组a排序，设枚举值为x，枚举区间[l,r),在r增加的情况下，要使区间内的数都大于a[r]-x，则l单调不递减，r单调递增。

附代码

#include<cstdio>
#include<cstring>
#include<string>
#include<algorithm>
#include<iostream>
#include<cmath>
#include<cstdlib>
#include<ctime>

using namespace std;
const int MAXN=100005;
int a[MAXN];

int main()
{
	int n;
	while (scanf("%d",&n)!=EOF)
	{
		int m=(n-1)*n/4;
		if ((n*(n-1)/2)&1)
		{
			m++;
		}
		if (n==1)
		{
			cout<<0<<endl;
			continue;
		}
		
		memset(a,0,sizeof(a));
		for (int i=1;i<=n;i++)
		{
			scanf("%d",&a[i]);
		}
		sort(a+1,a+1+n);
		long long l=0,r=a[n],mid,num,j;
		while (l<r)
		{
			mid=(l+r)>>1;
			j=1;
			num=0;
			for (int i=2;i<=n;i++)
			{
				while (a[i]-a[j]>mid)
				{
					j++;
				}
				num+=(i-j);
			}
			if (num>=m)
			{
				r=mid;
			}
			else
			{
				l=mid+1;
			}
		}
		cout<<l<<endl;
	}
    return 0;
}

哇昨天一天A了4个题