POJ3579 Median —— 二分

题目链接：http://poj.org/problem?id=3579

Median

Time Limit: 1000MS		Memory Limit: 65536K
Total Submissions: 8286		Accepted: 2892

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

 

 

 

 

题解：

1.对数组进行排序。计算出有多少对数，并计算出中位数所在的位置m。

2.二分中位数mid，然后检测有多少对数的差小于等于mid。假设有cnt对，如果cnt>=m，那么缩小中位数；否则扩大中位数。

3.注意：upper_bound()、lower_bound()的区间是前闭后开。

 

 

代码如下：

 #include <iostream>
 #include <cstdio>
 #include <cstring>
 #include <cmath>
 #include <algorithm>
 #include <vector>
 #include <queue>
 #include <stack>
 #include <map>
 #include <string>
 #include <set>
 #define ms(a,b) memset((a),(b),sizeof((a)))
 using namespace std;
 typedef long long LL;
 const double EPS = 1e-;
 const int INF = 2e9;
 const LL LNF = 2e18;
 const int MAXN = 1e5+;

 int n, a[MAXN];
 int m;

 bool test(int mid)
 {
     int cnt = ;
     for(int i = ; i<=n; i++)   //注意，对于每个数，只需往一边找，否则会出现重复计算。
         cnt += upper_bound(a+i+, a++n, a[i]+mid)-(a+i+); //在[i+1, n+1)的范围，即[i+1,n]
     return cnt>=m;
 }

 int main()
 {
     while(scanf("%d", &n)!=EOF)
     {
         for(int i = ; i<=n; i++)
             scanf("%d", &a[i]);

         sort(a+, a++n);
         m = n*(n-)/;  //有多少个数
         m = (m+)/;    //中位数所在的位置
         int l = , r = a[n]-a[];
         while(l<=r)
         {
             int mid = (l+r)>>;
             if(test(mid))
                 r = mid - ;
             else
                 l = mid + ;
         }
         printf("%d\n", l);
     }
 }