CF1012A Photo of The Sky(思考，模拟)
A. Photo of The Sky
time limit per test
1 second
memory limit per test
256 megabytes
input
standard input
output
standard output
Pavel made a photo of his favourite stars in the sky. His camera takes a photo of all points of the sky that belong to some rectangle with sides parallel to the coordinate axes.
Strictly speaking, it makes a photo of all points with coordinates (x,y)(x,y), such that x1≤x≤x2x1≤x≤x2 and y1≤y≤y2y1≤y≤y2, where (x1,y1)(x1,y1) and (x2,y2)(x2,y2) are coordinates of the left bottom and the right top corners of the rectangle being photographed. The area of this rectangle can be zero.
After taking the photo, Pavel wrote down coordinates of nn of his favourite stars which appeared in the photo. These points are not necessarily distinct, there can be multiple stars in the same point of the sky.
Pavel has lost his camera recently and wants to buy a similar one. Specifically, he wants to know the dimensions of the photo he took earlier. Unfortunately, the photo is also lost. His notes are also of not much help; numbers are written in random order all over his notepad, so it's impossible to tell which numbers specify coordinates of which points.
Pavel asked you to help him to determine what are the possible dimensions of the photo according to his notes. As there are multiple possible answers, find the dimensions with the minimal possible area of the rectangle.
Input
The first line of the input contains an only integer nn (1≤n≤1000001≤n≤100000), the number of points in Pavel's records.
The second line contains 2⋅n2⋅n integers a1a1, a2a2, ..., a2⋅na2⋅n (1≤ai≤1091≤ai≤109), coordinates, written by Pavel in some order.
Output
Print the only integer, the minimal area of the rectangle which could have contained all points from Pavel's records.
Examples
input
Copy
4 4 1 3 2 3 2 1 3
output
Copy
1
input
Copy
3 5 8 5 5 7 5
output
Copy
0
Note
In the first sample stars in Pavel's records can be (1,3)(1,3), (1,3)(1,3), (2,3)(2,3), (2,4)(2,4). In this case, the minimal area of the rectangle, which contains all these points is 11 (rectangle with corners at (1,3)(1,3) and (2,4)(2,4)).
给出一堆点的x和y（无序的，无法确定是x还是y）求能覆盖这些点的长方形的最小面积
对x,y进行排序后，得到一个递增序列，我们暂且将他当成这样：
最大值可能在两种情况取到
或是：
然后不断遍历i得出最小值即可
代码：
#include<iostream>
#include<cstdio>
#include<algorithm>
using namespace std;
const int INF=0x3f3f3f3f;
const int size=1e5+5;
int arr[2*size];
int main()
{
int n;
while(cin>>n)
{
int i;
for(i=0;i<2*n;i++)
{
scanf("%d",&arr[i]);
}
sort(arr,arr+2*n);
long long a=arr[n-1]-arr[0];
long long b=arr[2*n-1]-arr[n];
long long minn=a*b;
a=arr[2*n-1]-arr[0];
for(int i=1;i<n;i++)
{
b=arr[i+n-1]-arr[i];
minn=min(minn,a*b);
}
cout<<minn<<endl;
}
}
