Axis-Parallel Rectangle

D - Axis-Parallel Rectangle

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Time limit : 2sec / Memory limit : 256MB

Score : 400 points

Problem Statement

We have N points in a two-dimensional plane.
The coordinates of the i-th point (1≤i≤N) are (xi,yi).
Let us consider a rectangle whose sides are parallel to the coordinate axes that contains K or more of the N points in its interior.
Here, points on the sides of the rectangle are considered to be in the interior.
Find the minimum possible area of such a rectangle.

Constraints

• 2≤K≤N≤50
• −109≤xi,yi≤109(1≤i≤N)
• xi≠xj(1≤i<j≤N)
• yi≠yj(1≤i<j≤N)
• All input values are integers. (Added at 21:50 JST)

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Input

Input is given from Standard Input in the following format:

N K
x1 y1
:
xN yN

Output

Print the minimum possible area of a rectangle that satisfies the condition.

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Sample Input 1

Copy

4 4
1 4
3 3
6 2
8 1

Sample Output 1

Copy

21

One rectangle that satisfies the condition with the minimum possible area has the following vertices: (1,1), (8,1), (1,4) and (8,4).
Its area is (8−1)×(4−1)=21.

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Sample Input 2

Copy

4 2
0 0
1 1
2 2
3 3

Sample Output 2

Copy

1

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Sample Input 3

Copy

4 3
-1000000000 -1000000000
1000000000 1000000000
-999999999 999999999
999999999 -999999999

Sample Output 3

3999999996000000001

Watch out for integer overflows.

 

 

 

// N 个点，选出一个最小的矩形，包括至少 k 个点，求这最小的矩形的面积

// 枚举，疯狂枚举就行

 #include <bits/stdc++.h>
 using namespace std;
 #define MOD 998244353
 #define INF 0x3f3f3f3f3f3f3f3f
 #define LL long long
 #define MX 55
 struct Node
 {
     LL x, y;
     bool operator < (const Node &b)const{
         return x<b.x;
     }
 }pt[MX];

 int k, n;

 int main()
 {
     scanf("%d%d",&n,&k);
     for (int i=;i<=n;i++)
         scanf("%lld%lld",&pt[i].x, &pt[i].y);
     sort(pt+,pt++n);
     LL area = INF;
     for (int i=;i<=n;i++)
     {
         for (int j=i+;j<=n;j++)
         {
             LL miny = min(pt[i].y, pt[j].y);
             LL maxy = max(pt[i].y, pt[j].y);
             for (int q=;q<=n;q++)
             {
                 if (pt[q].y>maxy||pt[q].y<miny) continue;
                 int tot = ;
                 for (int z=q;z<=n;z++)
                 {
                     if (pt[z].y>maxy||pt[z].y<miny) continue;
                     tot++;
                     if (tot>=k)
                         area = min(area, (maxy-miny)*(pt[z].x-pt[q].x));
                 }
             }
         }
     }
     printf("%lld\n",area);
     return ;
 }