Question

我有一个代码，使用旋转卡尺算法定义了距离最长的两个点。

代码在第一行中获取点N的数量，然后N次获取点X，Y的坐标。之后显示最长距离的长度。

例如：

INPUT
6
1 1
-1 0
-3 -1
-2 -2
2 3
4 -2

OUTPUT
7.0710678119

INPUT
6
2 2
0 -3
5 7
3 3
2 1
-1 1

OUTPUT
4.47213595499958 #my comment: from (3,3) to (5,7)

但是在某些情况下， 3个或更多点位于一条直线上。然后我该怎么做？

from math import *

def rast(x1, x2, y1, y2):
        x = x2-x1
        y = y2-y1
        l = sqrt(pow(fabs(x), 2)+pow(fabs(y), 2));
        return l

def orientation(p,q,r):
    '''Return positive if p-q-r are clockwise, neg if ccw, zero if colinear.'''
    return (q[1]-p[1])*(r[0]-p[0]) - (q[0]-p[0])*(r[1]-p[1])

def hulls(Points):
    '''Graham scan to find upper and lower convex hulls of a set of 2d points.'''
    U = []
    L = []
    Points.sort()
    for p in Points:
        while len(U) > 1 and orientation(U[-2],U[-1],p) <= 0: U.pop()
        while len(L) > 1 and orientation(L[-2],L[-1],p) >= 0: L.pop()
        U.append(p)
        L.append(p)
    return U,L

def rotatingCalipers(Points):
    '''Given a list of 2d points, finds all ways of sandwiching the points
between two parallel lines that touch one point each, and yields the sequence
of pairs of points touched by each pair of lines.'''
    U,L = hulls(Points)
    i = 0
    j = len(L) - 1
    while i < len(U) - 1 or j > 0:
        yield U[i],L[j]

        # if all the way through one side of hull, advance the other side
        if i == len(U) - 1: j -= 1
        elif j == 0: i += 1

        # still points left on both lists, compare slopes of next hull edges
        # being careful to avoid divide-by-zero in slope calculation
        elif (U[i+1][1]-U[i][1])*(L[j][0]-L[j-1][0]) > \
                (L[j][1]-L[j-1][1])*(U[i+1][0]-U[i][0]):
            i += 1
        else: j -= 1

def diameter(Points):
    '''Given a list of 2d points, returns the pair that's farthest apart.'''
    diam,pair = max([((p[0]-q[0])**2 + (p[1]-q[1])**2, (p,q))
                     for p,q in rotatingCalipers(Points)])
    return pair

n=int(input())
dots = []
for i in range(n):
    tmp = [int(j) for j in input().split()]
    dots.append([tmp[0],tmp[1]])
tmp = diameter(dots)
d1,d2=tmp[0],tmp[1]
print(rast(d1[0],d2[0],d1[1],d2[1]))

Answer 1

一旦制作了凸包，就需要对线条进行最长排序。在我的示例中，您将有红线，然后是带箭头的蓝色。

取最长的线（红色）并获得其角度。对于凸包中的每个点，检查S和P之间的线是否等于角度。如果这样，则计算两条线SP和EP的距离，如果一条线比蓝色线（即最长的线）长，则可以停止。否则，忽略红线，并沿第二条最长的线走。当没有相等的角度时，您可以停下来。

数组中两点之间的最长路径

1 个答案: