POJ 3579 Median 二分加判断

Median

Time Limit: 1000MS		Memory Limit: 65536K
Total Submissions: 12453		Accepted: 4357

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

4
1 3 2 4
3
1 10 2

Sample Output

1
8

Source

POJ Founder Monthly Contest – 2008.04.13, Lei Tao

 

二分答案+O(nlogn)判断合法性

 

O(n^2)是明显不行的。

二分枚举中位数，在判断时候枚举每一个arr[i] 作为左端点，然后使用 upper_bound 求出第一个大于 arr[i]+mid 的元素的下标j。

(j-1)-i 就是以arr[i]作为较小的数，可能的方案数。这样判断合法性的复杂度就是 O(nlogn)。

 

btw，这里不光可以求出中位数，其实可以求出任意的第m个数。

 

#include<cstdio>
#include<algorithm>
using namespace std;
#define ll long long

const int maxn = 1e5+;
int n;
ll m;
int arr[maxn];
bool valid(int mid){
    int cnt = ;
    for(int i=;i<n;i++){
        cnt += (upper_bound(arr+i,arr+n,arr[i]+mid)--(arr+i));
    }
    return cnt >= m;
}
int main(){
    while(scanf("%d",&n)!=EOF){
        for(int i=;i<n;i++)
            scanf("%d",&arr[i]);
        sort(arr,arr+n);
        m = 1ll*n*(n-)/;
        if(m%==){
            m = m/;
        }else m = m/+;
        int l = -, r = arr[n-]-arr[];
        while(l <= r){
            int mid = (l+r)/;
            if(valid(mid))
                r = mid-;
            else
                l = mid+;
        }
        printf("%d\n",l);
    }
    return ;
}