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 個點，選出一個最小的矩形，包括至少 k 個點，求這最小的矩形的面積 // 枚舉，瘋狂枚舉就行 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

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轉(zhuǎn)載于:https://www.cnblogs.com/haoabcd2010/p/7673891.html

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