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最新推荐文章于 2019-09-07 17:17:12 发布

ElephantFlySong

最新推荐文章于 2019-09-07 17:17:12 发布

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分类专栏：凸包

本文链接：https://blog.youkuaiyun.com/qust1508060414/article/details/52752863

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本文介绍了一种基于计算几何的方法来求解凸包上两点间最小距离的问题，并提供了完整的C++实现代码。该方法首先对输入点进行排序并去除重复点，然后使用Graham扫描算法构建凸包，最后计算通过凸包的直线到所有点的重心的最小距离。

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题目链接：https://uva.onlinejudge.org/index.php?option=com_onlinejudge&Itemid=8&page=show_problem&problem=2109

就是刘汝佳训练指南上的思路，需要注意的是只有两个点的情况。

#include <iostream>
#include <cstdio>
#include <cmath>
#include <algorithm>
using namespace std;

struct Point
{
    double x, y;

    Point() {}
    Point(double x, double y):x(x), y(y) {}

    bool operator < (const Point& a) const
    {
        if (x == a.x)
            return y < a.y;
        return x < a.x;
    }

    bool operator == (const Point& a) const
    {
        return x == a.x && y == a.y;
    }
};
typedef Point Vector;

Point P[10010], convex[10010];

Vector operator - (Point A, Point B)
{
    return Vector(A.x-B.x, A.y-B.y);
}

double Cross(Vector A, Vector B)
{
    return A.x*B.y - A.y*B.x;
}

int ConvexHull(Point *p, int n, Point *ch)
{
    sort(p, p+n);
    n = unique(p, p+n) - p;

    int m = 0;
    for (int i=0; i<n; i++)
    {
        while (m > 1 && Cross(ch[m-1]-ch[m-2], p[i]-ch[m-2]) <= 0) m--;
        ch[m++] = p[i];
    }

    int k = m;
    for (int i=n-2; i>=0; i--)
    {
        while (m > k && Cross(ch[m-1]-ch[m-2], p[i]-ch[m-2]) <= 0) m--;
        ch[m++] = p[i];
    }

    return m;
}

int main()
{
    int t;

    scanf("%d",&t);
    for (int Case=1; Case <=t; Case++)
    {
        int n;
        double sum_x = 0, sum_y = 0;

        scanf("%d",&n);
        for (int i=0; i<n; i++)
        {
            scanf("%lf%lf",&P[i].x,&P[i].y);
            sum_x += P[i].x;
            sum_y += P[i].y;
        }

        int cnt = ConvexHull(P,n,convex);
        double ans = 100000000000;
        for (int i=0; i<cnt-1; i++)
        {
            double A = convex[i].y - convex[i+1].y;
            double B = convex[i+1].x - convex[i].x;
            double C = convex[i].x*convex[i+1].y - convex[i].y*convex[i+1].x;

            double temp = (A*sum_x + B*sum_y + n*C) / sqrt(A*A + B*B);
            ans = abs(temp) < ans ? abs(temp) : ans;
        }

        if (n <= 2)
            ans = 0;
        printf("Case #%d: %.3lf\n",Case, ans/n);
    }

    return 0;
}