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，该三角形的任意两条边位于坐标轴（X 轴和 Y 轴）上以覆盖所有点。
注意：如果点位于三角形内或三角形的边上，则该点被覆盖。
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。所以覆盖所有点的路径是坐标轴上的 (1, 4) 和 (4, 1)。
下面是解决这个问题的分步算法：

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“answer”中。
3. 用前一个总和替换点的下一个最大总和。
4. 然后，您将获得坐标轴 ( 1, answer ) 和 ( answer, 1 ) 上的路径，该坐标轴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
?>

Javascript

 
// Javascript program to illustrate
// the above problem
 
// function to get the minimum length of
// the shorter side of the triangle
function shortestLength(n, x, y)
{
    let answer = 0;
 
    // traversing through each
    // polets on the plane
    let i = 0;
    while (n != 0 && i < x.length)
    {
        // if sum of a polets is greater
        // than the previous one, the
        // maximum gets replaced
        if (x[i] + y[i] > answer)
            answer = x[i] + y[i];
 
        i++;
    }
 
    // prlet the length
    document.write("Length -> " + answer + "<br/>");
    document.write("Path -> " +
                       "( 1, " + answer + " )" +
                       "and ( " + answer + ", 1 )");
}
 
 
// Driver Code
 
    // initialize the number of polets
    let n = 4;
 
    // polets on the plane
    let x = [ 1, 4, 2, 1 ];
    let y = [ 4, 1, 1, 2 ];
 
    shortestLength(n, x, y);
 

输出：

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

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