Axis-Parallel Rectangle

D - Axis-Parallel Rectangle

---

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)

---

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.

---

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.

---

Sample Input 2

Copy

4 2
0 0
1 1
2 2
3 3

Sample Output 2

Copy

1

---

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=1;i<=n;i++)
        scanf("%lld%lld",&pt[i].x, &pt[i].y);
    sort(pt+1,pt+1+n);
    LL area = INF;
    for (int i=1;i<=n;i++)
    {
        for (int j=i+1;j<=n;j++)
        {
            LL miny = min(pt[i].y, pt[j].y);
            LL maxy = max(pt[i].y, pt[j].y);
            for (int q=1;q<=n;q++)
            {
                if (pt[q].y>maxy||pt[q].y<miny) continue;
                int tot = 0;
                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
",area);
    return 0;
}