poj3579 双重二分，有意思的一道题，注意二分边界

Median

Time Limit: 1000MS		Memory Limit: 65536K
Total Submissions: 5246		Accepted: 1686

Description

Given N numbers, X1,
X2, ... , XN, let us calculate the difference of every pair of numbers: ∣Xi- Xj∣ (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, ... ,
XN, ( Xi
≤ 1,000,000,000  3 ≤ N ≤ 1,00,000 )

Output

For each test case, output the m#include<iostream>
#include<algorithm>
#include<cstdio>
using namespace std;
int a[100005];
int n,num;
int bin(int x){
//cout<<"x:"<<x<<endl;
int l=0,r=n-1,mid;
while(l<r){
mid=(l+r)/2;
if (a[mid]<=x){
l=mid+1;
}
else r=mid;
}
//cout<<"bin:"<<n-1-r+1<<endl;
return n-1-r+1;
}
int pan(int guess){
//cout<<"guess:"<<guess<<endl;
int count=0;
for (int i=0;i<n;i++){
if (a[i]+guess>=a[n-1]) break;
count+=bin(a[i]+guess);
if (count>num) return -1;
}
if (count<=num) return 0;
}
int main(){
while(scanf("%d",&n)!=EOF){
num=(n)*(n-1)/2;
num/=2;
for (int i=0;i<n;i++){
scanf("%d",&a[i]);
}
sort(a,a+n);

4000
int r=a[n-1]-a[0];
int l=0,mid;
int flag;
while(l<r){
mid=(l+r)/2;
flag=pan(mid);
if (flag==0){
r=mid;
}
else if (flag==-1) l=mid+1;
}
printf("%d\n",l);
}
}

edian in a separate line.

Sample Input

4
1 3 2 4
3
1 10 2

Sample Output

1
8

排序之后，从0 到 最大差值进行二分搜索，判断guess值是否满足大于a[i]+guess的一共有count个，这里计数用到了第二次二分，找到上界，返回末尾与它的下标2差，如果没有那么多，说明guess值猜大了，有的话还是要继续猜，因为不能保证当前guess值是在原数组中出现过的，