三角形最短路径的最小长度

给定平面上的N个点，(X ₁ ，Y ₁ )，(X ₂ ，Y ₂ )，(X ₃ ，Y ₃ )，……，(X _N ，Y _N )。任务是计算三角形较短边的最小长度。以及放置等腰三角形的路径或点，该三角形的任意两个边都位于坐标轴(X轴和Y轴)上，以覆盖所有点。

注意：如果一个点位于三角形内部或三角形侧面，则将覆盖该点。

例子：

Input: (1, 3), (1, 1), (2, 1), (2, 2)
Output: Length -> 4 , Path -> ( 1, 4 ) and ( 4, 1 )

Input: (1, 2), (1, 1), (2, 1)
Output: Length -> 3 , Path -> ( 1, 3 ) and ( 3, 1 )

在第一个示例中，最短路径的最小长度等于点的最大和，即1 + 3或2 + 2。因此，将覆盖所有点的路径在坐标轴上分别为(1、4)和(4、1)。

以下是解决此问题的分步算法：

1. 在平面上初始化“ N”个点。
2. 遍历每个点并找到每个点的总和并将其存储在变量“答案”中。
3. 用先前的总和替换下一个最大的总和。
4. 然后，您将获得坐标轴(1，answer)和(answer，1)上的路径，该路径将覆盖等腰三角形的所有点。

下面是上述算法的实现：

C++

// C++ program to illustrate
// the above problem
#include 
using namespace std;
#define ll long long
  
// function to get the minimum length of
// the shorter side of the triangle
void shortestLength(int n, int x[], int y[])
{
        int answer = 0;
  
    // traversing through each points on the plane
    int i = 0;
    while (n--) {
        // if sum of a points is greater than the
        // previous one, the maximum gets replaced
        if (x[i] + y[i] > answer)
            answer = x[i] + y[i];
  
        i++;
    }
  
    // print the length
    cout << "Length -> " << answer << endl;
    cout << "Path -> "
         << "( 1, " << answer << " )"
         << "and ( " << answer << ", 1 )";
}
  
// Driver code
int main()
{
    // initialize the number of points
    int n = 4;
  
    // points on the plane
    int x[n] = { 1, 4, 2, 1 };
    int y[n] = { 4, 1, 1, 2 };
  
    shortestLength(n, x, y);
  
    return 0;
}

Java

// Java program to illustrate 
// the above problem 
class GFG
{
  
// function to get the minimum length of 
// the shorter side of the triangle 
static void shortestLength(int n, int x[], 
                           int y[]) 
{ 
    int answer = 0; 
  
    // traversing through each 
    // points on the plane 
    int i = 0; 
    while (n != 0 && i < x.length)
    { 
        // if sum of a points is greater
        // than the previous one, the 
        // maximum gets replaced 
        if (x[i] + y[i] > answer) 
            answer = x[i] + y[i]; 
  
        i++; 
    } 
  
    // print the length 
    System.out.println("Length -> " + answer ); 
    System.out.println("Path -> " + 
                       "( 1, " + answer + " )" + 
                       "and ( " + answer + ", 1 )"); 
} 
  
// Driver code 
public static void main(String[] args) 
{ 
    // initialize the number of points 
    int n = 4; 
  
    // points on the plane 
    int x[] = new int[] { 1, 4, 2, 1 }; 
    int y[] = new int[] { 4, 1, 1, 2 }; 
  
    shortestLength(n, x, y); 
} 
}
  
// This code is contributed 
// by Prerna Saini

Python 3

# Python 3 program to illustrate
# the above problem
  
# function to get the minimum length of
# the shorter side of the triangle
def shortestLength(n, x, y):
    answer = 0
  
    # traversing through each
    # points on the plane
    i = 0
    while n > 0: 
          
        # if sum of a points is greater 
        # than the previous one, the 
        # maximum gets replaced
        if (x[i] + y[i] > answer):
            answer = x[i] + y[i]
  
        i += 1
        n -= 1
  
    # print the length
    print("Length -> "+ str(answer))
    print( "Path -> "+
        "( 1, " +str(answer)+ " )"+
        "and ( "+str( answer) +", 1 )")
  
# Driver code
if __name__ == "__main__":
      
    # initialize the number of points
    n = 4
  
    # points on the plane
    x = [ 1, 4, 2, 1 ]
    y = [ 4, 1, 1, 2 ] 
    shortestLength(n, x, y)
  
# This code is contributed 
# by ChitraNayal

C#

// C# program to illustrate 
// the above problem 
using System;
  
class GFG
{
  
// function to get the minimum 
// length of the shorter side 
// of the triangle 
static void shortestLength(int n, int[] x, 
                           int[] y) 
{ 
    int answer = 0; 
  
    // traversing through each 
    // points on the plane 
    int i = 0; 
    while (n != 0 && i < x.Length)
    { 
        // if sum of a points is greater
        // than the previous one, the 
        // maximum gets replaced 
        if (x[i] + y[i] > answer) 
            answer = x[i] + y[i]; 
  
        i++; 
    } 
  
    // print the length 
    Console.WriteLine("Length -> " + answer); 
    Console.WriteLine("Path -> " + 
                      "( 1, " + answer + " )" + 
                      "and ( " + answer + ", 1 )"); 
} 
  
// Driver code
static public void Main ()
{
      
    // initialize the 
    // number of points 
    int n = 4; 
  
    // points on the plane 
    int[] x = new int[] { 1, 4, 2, 1 }; 
    int[] y = new int[] { 4, 1, 1, 2 }; 
  
    shortestLength(n, x, y); 
}
}
  
// This code is contributed by Mahadev

PHP

 $answer)
            $answer = $x[$i] + $y[$i];
  
        $i++;
    }
  
    // print the length
    echo "Length -> ".$answer."\n";
    echo "Path -> ". "( 1, " .$answer ." )".
         "and ( " .$answer . ", 1 )";
}
  
// Driver code
  
// initialize the number of points
$n = 4;
  
// points on the plane
$x = array(1, 4, 2, 1 );
$y = array(4, 1, 1, 2 );
  
shortestLength($n, $x, $y);
  
// This code is contributed 
// by ChitraNayal
?>

输出：

Length -> 5
Path -> ( 1, 5 )and ( 5, 1 )