Axis-Parallel Rectangle

最新推荐文章于 2021-05-23 18:11:18 发布

转载最新推荐文章于 2021-05-23 18:11:18 发布 · 378 阅读

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原文链接：http://www.cnblogs.com/haoabcd2010/p/7673891.html

本文介绍了一个关于在二维平面上寻找包含至少K个点的最小面积矩形的问题，并提供了一种通过枚举来解决该问题的方法。输入为N个点坐标，输出是最小矩形的面积。

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

Sample Output 1

Copy

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

Sample Output 2

Copy

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

 1 #include <bits/stdc++.h>
 2 using namespace std;
 3 #define MOD 998244353
 4 #define INF 0x3f3f3f3f3f3f3f3f
 5 #define LL long long
 6 #define MX 55
 7 struct Node
 8 {
 9     LL x, y;
10     bool operator < (const Node &b)const{
11         return x<b.x;
12     }
13 }pt[MX];
14 
15 int k, n;
16 
17 int main()
18 {
19     scanf("%d%d",&n,&k);
20     for (int i=1;i<=n;i++)
21         scanf("%lld%lld",&pt[i].x, &pt[i].y);
22     sort(pt+1,pt+1+n);
23     LL area = INF;
24     for (int i=1;i<=n;i++)
25     {
26         for (int j=i+1;j<=n;j++)
27         {
28             LL miny = min(pt[i].y, pt[j].y);
29             LL maxy = max(pt[i].y, pt[j].y);
30             for (int q=1;q<=n;q++)
31             {
32                 if (pt[q].y>maxy||pt[q].y<miny) continue;
33                 int tot = 0;
34                 for (int z=q;z<=n;z++)
35                 {
36                     if (pt[z].y>maxy||pt[z].y<miny) continue;
37                     tot++;
38                     if (tot>=k)
39                         area = min(area, (maxy-miny)*(pt[z].x-pt[q].x));
40                 }
41             }
42         }
43     }
44     printf("%lld\n",area);
45     return 0;
46 }

View Code

转载于:https://www.cnblogs.com/haoabcd2010/p/7673891.html