POJ 2836 Rectangular Covering

最新推荐文章于 2019-05-07 22:15:24 发布

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本文探讨了如何使用一组边平行于坐标轴的矩形覆盖平面上的n个点，并确保每个点至少被一个矩形覆盖的问题。目标是最小化所有矩形的总面积。文章提供了一种基于动态规划的方法来解决此问题。

Rectangular Covering

Time Limit: 1000MS		Memory Limit: 65536K
Total Submissions: 2813		Accepted: 800

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.

#include<cstdio>
#include<string.h>
#include<algorithm>
using namespace std;
int x[17],y[17];
int dp[1<<16];
struct node
{
    int area,S;
}c[404];
int main()
{
    int n;
    while(~scanf("%d",&n)&&n)
    {
        for(int i=0;i<n;i++)
            scanf("%d%d",&x[i],&y[i]);
        int tol=0;
        for(int i=0;i<n;i++)
        {
            for(int j=i+1;j<n;j++)//枚举每一种组合的两个端点
            {
                c[tol].area=max(1,abs(x[i]-x[j]))*max(1,abs(y[i]-y[j]));
                c[tol].S=(1<<i)|(1<<j);
                for(int k=0;k<n;k++)//在里面的点
                {
                    if((x[k]-x[i])*(x[k]-x[j])<=0&&(y[k]-y[i])*(y[k]-y[j])<=0)
                        c[tol].S=c[tol].S|(1<<k);
                }
                tol++;
            }
        }
        for(int i=0;i<1<<n;i++)dp[i]=1e9;
        dp[0]=0;
        for(int S=0;S<1<<n;S++)//dp
        {
            for(int i=0;i<tol;i++)
            {
                int p=S|c[i].S;
                dp[p]=min(dp[p],dp[S]+c[i].area);
            }
        }
        printf("%d\n",dp[(1<<n)-1]);
    }
    return 0;
}