给定 xy 坐标平面上的三个点。你需要找到没有。通过使多段线通过这些点形成的线段。 （线段可以垂直或水平对齐坐标轴）
例子 ：
Input : A = {-1, -1}, B = {-1, 3}, C = {4, 3}
Output : 2
Expantaion:
There are two segments in this polyline.
Input :A = {1, 1}, B = {2, 3} C = {3, 2}
Output : 3
如果所有点都在 x 轴或 y 轴上，则结果为 1。如果点可以形成 L 形，则结果为 2。如果三个点中的任何一个可以用作连接点，则形成 L 形。否则答案为 3。
C++
// CPP program to find number of horizontal (or vertical
// line segments needed to connect three points.
#include
using namespace std;
// Function to check if the third point forms a
// rectangle with other two points at corners
bool isBetween(int a, int b, int c)
{
return min(a, b) <= c && c <= max(a, b);
}
// Returns true if point k can be used as a joining
// point to connect using two line segments
bool canJoin(int x[], int y[], int i, int j, int k)
{
// Check for the valid polyline with two segments
return (x[k] == x[i] || x[k] == x[j]) &&
isBetween(y[i], y[j], y[k]) ||
(y[k] == y[i] || y[k] == y[j]) &&
isBetween(x[i], x[j], x[k]);
}
int countLineSegments(int x[], int y[])
{
// Check whether the X-coordinates or
// Y-cocordinates are same.
if ((x[0] == x[1] && x[1] == x[2]) ||
(y[0] == y[1] && y[1] == y[2]))
return 1;
// Iterate over all pairs to check for two
// line segments
else if (canJoin(x, y, 0, 1, 2) ||
canJoin(x, y, 0, 2, 1) ||
canJoin(x, y, 1, 2, 0))
return 2;
// Otherwise answer is three.
else
return 3;
}
// Driver code
int main()
{
int x[3], y[3];
x[0] = -1; y[0] = -1;
x[1] = -1; y[1] = 3;
x[2] = 4; y[2] = 3;
cout << countLineSegments(x, y);
return 0;
} Java
// Java program to find number of horizontal
// (or vertical line segments needed to
// connect three points.
import java.io.*;
class GFG {
// Function to check if the third
// point forms a rectangle with
// other two points at corners
static boolean isBetween(int a, int b, int c)
{
return (Math.min(a, b) <= c &&
c <= Math.max(a, b));
}
// Returns true if point k can be
// used as a joining point to connect
// using two line segments
static boolean canJoin(int x[], int y[],
int i, int j, int k)
{
// Check for the valid polyline
// with two segments
return (x[k] == x[i] || x[k] == x[j]) &&
isBetween(y[i], y[j], y[k]) ||
(y[k] == y[i] || y[k] == y[j]) &&
isBetween(x[i], x[j], x[k]);
}
static int countLineSegments(int x[], int y[])
{
// Check whether the X-coordinates or
// Y-cocordinates are same.
if ((x[0] == x[1] && x[1] == x[2]) ||
(y[0] == y[1] && y[1] == y[2]))
return 1;
// Iterate over all pairs to check for two
// line segments
else if (canJoin(x, y, 0, 1, 2) ||
canJoin(x, y, 0, 2, 1) ||
canJoin(x, y, 1, 2, 0))
return 2;
// Otherwise answer is three.
else
return 3;
}
// Driver code
public static void main (String[] args) {
int x[]=new int[3], y[]=new int[3];
x[0] = -1; y[0] = -1;
x[1] = -1; y[1] = 3;
x[2] = 4; y[2] = 3;
System.out.println(countLineSegments(x, y));
}
}
// This code is contributed by vt_mPython3
# Python program to find number
# of horizontal (or vertical
# line segments needed to
# connect three points.
import math
# Function to check if the
# third point forms a
# rectangle with other
# two points at corners
def isBetween(a, b, c) :
return min(a, b) <= c and c <= max(a, b)
# Returns true if point k
# can be used as a joining
# point to connect using
# two line segments
def canJoin( x, y, i, j, k) :
# Check for the valid polyline
# with two segments
return (x[k] == x[i] or x[k] == x[j]) and isBetween(y[i], y[j], y[k]) or (y[k] == y[i] or y[k] == y[j]) and isBetween(x[i], x[j], x[k])
def countLineSegments( x, y):
# Check whether the X-coordinates or
# Y-cocordinates are same.
if ((x[0] == x[1] and x[1] == x[2]) or
(y[0] == y[1] and y[1] == y[2])):
return 1
# Iterate over all pairs to check for two
# line segments
elif (canJoin(x, y, 0, 1, 2) or
canJoin(x, y, 0, 2, 1) or
canJoin(x, y, 1, 2, 0)):
return 2
# Otherwise answer is three.
else:
return 3
#driver code
x= [-1,-1, 4]
y= [-1, 3, 3]
print(countLineSegments(x, y))
# This code is contributed by Gitanjali.C#
// C# program to find number of horizontal
// (or vertical) line segments needed to
// connect three points.
using System;
class GFG {
// Function to check if the third
// point forms a rectangle with
// other two points at corners
static bool isBetween(int a, int b, int c)
{
return (Math.Min(a, b) <= c &&
c <= Math.Max(a, b));
}
// Returns true if point k can be
// used as a joining point to connect
// using two line segments
static bool canJoin(int[] x, int[] y,
int i, int j, int k)
{
// Check for the valid polyline
// with two segments
return (x[k] == x[i] || x[k] == x[j])
&& isBetween(y[i], y[j], y[k])
|| (y[k] == y[i] || y[k] == y[j])
&& isBetween(x[i], x[j], x[k]);
}
static int countLineSegments(int[] x, int[] y)
{
// Check whether the X-coordinates or
// Y-cocordinates are same.
if ((x[0] == x[1] && x[1] == x[2]) ||
(y[0] == y[1] && y[1] == y[2]))
return 1;
// Iterate over all pairs to check for two
// line segments
else if (canJoin(x, y, 0, 1, 2)
|| canJoin(x, y, 0, 2, 1)
|| canJoin(x, y, 1, 2, 0))
return 2;
// Otherwise answer is three.
else
return 3;
}
// Driver code
public static void Main()
{
int[] x = new int[3];
int[] y = new int[3];
x[0] = -1;
y[0] = -1;
x[1] = -1;
y[1] = 3;
x[2] = 4;
y[2] = 3;
Console.WriteLine(countLineSegments(x, y));
}
}
// This code is contributed by vt_m.PHP
Javascript
输出 ：
2