POJ 2836 Rectangular Covering

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#动态规划

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本文探讨了如何通过使用一系列轴平行的矩形来覆盖平面上的给定点集，并确保每个点至少被一个矩形覆盖的同时，使得所有矩形总面积最小。文章采用状态压缩动态规划算法解决此问题。

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Description

n points are given on the Cartesian plane. Now you have to use some rectangles whose sides are parallel to the axes to cover them. Every point must be covered. And a point can be covered by several rectangles. Each rectangle should cover at least two points including those that fall on its border. Rectangles should have integral dimensions. Degenerate cases (rectangles with zero area) are not allowed. How will you choose the rectangles so as to minimize the total area of them?

Input

The input consists of several test cases. Each test cases begins with a line containing a single integer n (2 ≤ n ≤ 15). Each of the next n lines contains two integers x, y (−1,000 ≤ x, y ≤ 1,000) giving the coordinates of a point. It is assumed that no two points are the same as each other. A single zero follows the last test case.

Output

Output the minimum total area of rectangles on a separate line for each test case.

Sample Input

Sample Output

Hint

The total area is calculated by adding up the areas of rectangles used.

n<=15,想到状压dp。

将包含的点集状压，维护面积。

注意，矩形面积不能为零，一个矩形可能覆盖多个点。

#include<iostream>
#include<cstring>
#include<cstdio>
using namespace std;
const int N=16;
const int inf=1e9+7;
int n,x[N],y[N],s[N*N],w[N*N],tp,f[40005];
int ab(int x)
{
	return x<0?-x:x;
}
int main()
{
	while(scanf("%d",&n)&&n)
	{
		tp=0;
		for(int i=1;i<=n;i++)
			scanf("%d%d",&x[i],&y[i]);
		for(int i=1;i<=n;i++)
			for(int j=i+1;j<=n;j++)
			{
				++tp;
				int a,b;
				a=max(1,ab(x[i]-x[j])),b=max(1,ab(y[i]-y[j]));
				s[tp]=a*b;
				w[tp]=0;
				for(int k=1;k<=n;k++)
					if(x[k]>=min(x[i],x[j])&&x[k]<=max(x[i],x[j])&&y[k]>=min(y[i],y[j])&&y[k]<=max(y[i],y[j]))
						w[tp]|=(1<<k-1);
			}
		memset(f,0x3f,sizeof(f));
		f[0]=0;
		int end=(1<<n)-1;
		for(int i=0;i<=end;i++)
			if(f[i]<=1e9+7)
				for(int j=1;j<=tp;j++)
					if((i|w[j])!=i)
						f[i|w[j]]=min(f[i|w[j]],f[i]+s[j]);
		printf("%d\n",f[end]);
	}
	return 0;
}