POJ3579-Median

最新推荐文章于 2020-03-10 15:16:32 发布

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最新推荐文章于 2020-03-10 15:16:32 发布

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本文介绍了一种高效算法，用于解决给定一系列整数后，计算所有数对间差值绝对值构成的新数列中位数的问题。采用二分查找优化计算过程，显著提升了求解速度。

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Median

Time Limit: 1000MS		Memory Limit: 65536K
Total Submissions: 7377		Accepted: 2511

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

题意：给出一个数列，然后计算数列里各个数之间的差值的绝对值，形成一个新数列，求新数列的中位数

解题思路：算出新数列一共有多少个数，然后二分中间值，验证的时候也要通过二分来查找数字的位置

#include <iostream>
#include <cstdio>
#include <cstring>
#include <string>
#include <algorithm>
#include <queue>
#include <cmath>
#include <map>
#include <bitset>
#include <set>
#include <vector>
#include <functional>

using namespace std;

#define LL long long
const int INF=0x3f3f3f3f;

int a[100009],n,k;

bool check(int x)
{
    int sum=0;
    for(int i=1;i<n;i++)
    {
        int kk=upper_bound(a+i+1,a+1+n,x+a[i])-a-1;
        sum+=(kk-i);
    }
    if(sum>=k) return 1;
    else return 0;
}

int main()
{
    while(~scanf("%d",&n))
    {
        for(int i=1;i<=n;i++) scanf("%d",&a[i]);
        sort(a+1,a+1+n);
        k=n*(n-1)/2;
        if(k&1) k=(k+1)/2;
        else k/=2;
        int l=0,r=INF,ans;
        while(l<=r)
        {
            int mid=(l+r)>>1;
            if(check(mid)) {r=mid-1;ans=mid;}
            else l=mid+1;
        }
        printf("%d\n",ans);
    }
    return 0;
}