POJ 3579：Median 差值的中位数

Median

Time Limit: 1000MS		Memory Limit: 65536K
Total Submissions: 4680		Accepted: 1452

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

题意是给出一个数组，然后这些数组元素每一对之间都有一个差值，找出这些差值的中位数。

两次二分，第一次二分是枚举答案，第二次二分是我觉得很好玩的地方，在给定的差值下，找到多少个对的差值小于等于它，用这样的二分来判断这个差值在所有的差值中的位置。感觉很巧妙~

代码：

#include <iostream>

#include <algorithm>

#include <cmath>

#include <vector>

#include <string>

#include <cstring>

#pragma warning(disable:4996)

using namespace std;

int num,m;

int val[100005];

bool check(int mid)

{

	int i,sum=0;

	for (i = 0; i < num; i++)

	{

		sum += (lower_bound(val + i, val + num, val[i] + mid + 1) - (val + i) - 1);

	}

	if (sum >= m)

	{

		return true;

	}

	else

	{

		return false;

	}

}

int main()

{

	//freopen("i.txt","r",stdin);

	//freopen("o.txt","w",stdout);

	int i,left,right,mid;

	while (scanf("%d", &num) != EOF)

	{

		for (i = 0; i < num; i++)

		{

			scanf("%d", &val[i]);

		}

		m = (num*(num - 1) / 2 + 1) / 2;

		sort(val,val+num);

		left = 0;

		right = val[num - 1] - val[0];

		while (right-left > 1)

		{

			mid = (left + right) / 2;

			if (check(mid))

			{

				right = mid ;

			}

			else

			{

				left = mid ;

			}

		}

		cout << right << endl;

	}

	return 0;

}

巴特西

POJ 3579：Median 差值的中位数

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