CFgym：Glyph Recognition（计算几何）

最新推荐文章于 2024-10-30 00:13:06 发布

junior19

最新推荐文章于 2024-10-30 00:13:06 发布

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分类专栏：计算几何

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本文介绍了一种算法，该算法使用正3到8边形覆盖平面内的指定点集，并最大化内外多边形面积之比。通过转换点到极坐标并利用三角函数计算所需距离，最终确定最佳覆盖方案。

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▶ Given some points in the plane, and 3 ≤ k ≤ 8, find two regular k-gons containing all the points between them.

▶ Each polygon is centered at the origin, i.e. all vertices have equal distance to (0, 0).

▶ Each polygon has a vertex on the positive x-axis.

▶ The inner polygon is the largest such polygon containing none of the sample points.

▶ The outer polygon is the smallest such polygon containing all of the sample points

▶ Over all values of k, maximize the ratio Sinner/Souter of polygon areas.

题意：给一堆点，要求用正3~8边形任意一种，去覆盖这些点，要满足上述条件，令内正多边形面积/外正多边形面积最大。

思路：

要求的是所有点的OC长度，先将点化成极坐标，根据对称，将所有点转移到第一份格子即[0,360/K]内。知道OA的长度和角度，从而根据三角函数轻易求出OB和OC的长度，取长度最小和最大的比一下即可。

# include <bits/stdc++.h>
# define pi acos(-1.0)
using namespace std;
struct node
{
    double dis, ang;
}a[1003];
double cal(int k, int n)
{
    double f = 360.0/k*pi/180.0;
    double imax = 0, imin = 1e18, tmp;
    for(int i=0; i<n; ++i)
    {
        tmp = a[i].ang-(int)(a[i].ang/(360.0/k))*(360.0/k);
        tmp = tmp*pi/180.0;
        double len = cos(tmp-f/2)*(a[i].dis);
        len = len/cos(f/2.0);
        imax = max(imax, len);
        imin = min(imin, len);
    }
    return imin*imin/(imax*imax);
}
int main()
{
    int n, id;
    double x, y, ans = 0;
    scanf("%d",&n);
    for(int i=0; i<n; ++i)
    {
        scanf("%lf%lf",&x,&y);
        a[i].dis = sqrt(x*x+y*y);
        a[i].ang = atan(y/x)*180/pi;
        if(x < 0) a[i].ang += 180;
        if(a[i].ang < 0 && x >= 0) a[i].ang += 360;
    }
    for(int i=3; i<=8; ++i)
    {
        if(cal(i,n) > ans)
        {
            ans = cal(i, n);
            id = i;
        }
    }
    printf("%d %.10f\n",id,ans);
    return 0;
}