给定的两点表示的线和 .任务是找到可以通过单个点的最大线数,而不叠加(或覆盖)任何其他线。我们可以移动任何线,但不能旋转它。
例子:
Input : Line 1 : x1 = 1, y1 = 1, x2 = 2, y2 = 2
Line 2 : x2 = 2, y1 = 2, x2 = 4, y2 = 10
Output : 2
There are two lines. These two lines are not
parallel, so both of them will pass through
a single point.
Input : Line 1 : x1 = 1, y1 = 5, x2 = 1, y2 = 10
Line 2 : x2 = 5, y1 = 1, x2 = 10, y2 = 1
Output : 2
- 将线表示为对其中 line 可以表示为 ,称为线斜率形式。我们现在可以看到我们可以更改任何行的c ,但不能修改m 。
- 假定(c1 ≠ c2)具有相同m值的线平行。此外,任何两条平行线都不能通过同一点而不相互叠加。
- 因此,我们的问题简化为从给定的一组线中找到不同的斜率值。
我们可以计算一条线的斜率 ,将它们添加到集合中并计算集合中斜率的不同值的数量。但是我们必须单独处理垂直线。
因此,如果然后,斜率 = INT_MAX 。
否则,斜率 = .
下面是该方法的实现。
C++
// C++ program to find maximum number of lines
// which can pass through a single point
#include
using namespace std;
// function to find maximum lines which passes
// through a single point
int maxLines(int n, int x1[], int y1[],
int x2[], int y2[])
{
unordered_set s;
double slope;
for (int i = 0; i < n; ++i) {
if (x1[i] == x2[i])
slope = INT_MAX;
else
slope = (y2[i] - y1[i]) * 1.0
/ (x2[i] - x1[i]) * 1.0;
s.insert(slope);
}
return s.size();
}
// Driver program
int main()
{
int n = 2, x1[] = { 1, 2 }, y1[] = { 1, 2 },
x2[] = { 2, 4 }, y2[] = { 2, 10 };
cout << maxLines(n, x1, y1, x2, y2);
return 0;
}
// This code is written by
// Sanjit_Prasad
Java
// Java program to find maximum number of lines
// which can pass through a single point
import java.util.*;
import java.lang.*;
import java.io.*;
class GFG{
// function to find maximum lines which passes
// through a single point
static int maxLines(int n, int x1[], int y1[],
int x2[], int y2[])
{
Set s=new HashSet();
double slope;
for (int i = 0; i < n; ++i) {
if (x1[i] == x2[i])
slope = Integer.MAX_VALUE;
else
slope = (y2[i] - y1[i]) * 1.0
/ (x2[i] - x1[i]) * 1.0;
s.add(slope);
}
return s.size();
}
// Driver program
public static void main(String args[])
{
int n = 2, x1[] = { 1, 2 }, y1[] = { 1, 2 },
x2[] = { 2, 4 }, y2[] = { 2, 10 };
System.out.print(maxLines(n, x1, y1, x2, y2));
}
}
// This code is written by
// Subhadeep
Python3
# Python3 program to find maximum number
# of lines which can pass through a
# single point
import sys
# function to find maximum lines
# which passes through a single point
def maxLines(n, x1, y1, x2, y2):
s = [];
slope=sys.maxsize;
for i in range(n):
if (x1[i] == x2[i]):
slope = sys.maxsize;
else:
slope = (y2[i] - y1[i]) * 1.0 /(x2[i] - x1[i]) * 1.0;
s.append(slope);
return len(s);
# Driver Code
n = 2;
x1 = [ 1, 2 ];
y1 = [1, 2];
x2 = [2, 4];
y2 = [2, 10];
print(maxLines(n, x1, y1, x2, y2));
# This code is contributed by mits
C#
// C# program to find maximum number of lines
// which can pass through a single point
using System;
using System.Collections.Generic;
class GFG
{
// function to find maximum lines which passes
// through a single point
static int maxLines(int n, int []x1, int []y1,
int []x2, int []y2)
{
HashSet s = new HashSet();
double slope;
for (int i = 0; i < n; ++i)
{
if (x1[i] == x2[i])
slope = int.MaxValue;
else
slope = (y2[i] - y1[i]) * 1.0
/ (x2[i] - x1[i]) * 1.0;
s.Add(slope);
}
return s.Count;
}
// Driver code
public static void Main()
{
int n = 2;
int []x1 = { 1, 2 }; int []y1 = { 1, 2 };
int []x2 = { 2, 4 }; int []y2 = { 2, 10 };
Console.Write(maxLines(n, x1, y1, x2, y2));
}
}
/* This code contributed by PrinciRaj1992 */
PHP
Javascript
输出:
2
时间复杂度:
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