点 (X, Y) 处的碰撞次数

给定一个大小为N * 3的矩阵arr[][] ，其中每一行都包含定义一个元素的 3 个属性，即(x-coordinate, y-coordinate, speed)和两个整数X和Y ，任务是计算点(X, Y)处可能的碰撞对的数量，如果给定数组中的每个元素都以各自的速度沿直线向给定点(X, Y) 移动。
例子：

Input: arr[] = [ [5, 12, 1], [16, 63, 5], [-10, 24, 2], [7, 24, 2], [-24, 7, 2] ], X = 0, Y = 0 
Output: 4 
Explanation: 
Possible collisions are between the following pair of elements: (arr[0], arr[1]), (arr[0], arr[2]), (arr[1], arr[2]) and (arr[3], arr[4]) 
Hence, total collisions are 4.
Input: arr[] = [ [1, 42, 9] ], X = 1, Y = 1 
Output: 0 
Explanation: 
Since a single point is present, there will be no collisions. 
Hence, total collisions are 0.

方法：我们的想法是简单地找到一条其每一点将达到给定的点（X，Y）的时间。如果两点同时到达原点，那么它们肯定会发生碰撞。现在，要计算每个点所花费的时间，请执行以下操作：

For a given point (x, y) and speed S 
Distance D is given by: 
D = = √ ((x2 – x1)² + (y2 – y1)²) = √ ((x)² + (y)²) 
Time = Distance/Speed, squaring on both sides to eliminate root from distance, 
Time² = Distance²/Speed²

在为每个点计算时间²后，只需检查它们中的哪些相同并计算碰撞次数。
以下是步骤：

• 遍历数组arr[] 。
• 创建一个新数组Time[]并为每个点计算时间并附加到该数组并对其进行排序。
• Traverse Time[]并计算同时到达给定点的元素数量，使用该计数计算当时的碰撞次数。
• 最后，返回碰撞的总数。

下面是上述方法的实现：

C++14

// C++14 program to implement
// the above approach
#include 
using namespace std;
 
// Function to find the count of
// possible pairs of collisions
int solve(vector> &D, int N,
                           int X, int Y)
{
     
    // Stores the time at which
    // points reach the origin
    vector T;
 
    // Calculate time for each point
    for(int i = 0; i < N; i++)
    {
        int x = D[i][0];
        int y = D[i][1];
 
        double speed = D[i][2];
 
        double time = ((x * x - X * X) +
                       (y * y - Y * Y)) /
                       (speed * speed);
 
        T.push_back(time);
    }
 
    // Sort the times
    sort(T.begin(), T.end());
 
    int i = 0;
    int total = 0;
 
    // Counting total collisions
    while (i < T.size() - 1)
    {
         
        // Count of elements arriving at
        // a given point at the same time
        int count = 1;
 
        while (i < T.size() - 1 and
                T[i] == T[i + 1])
        {
            count += 1;
            i += 1;
        }
 
        total += (count * (count - 1)) / 2;
        i += 1;
    }
    return total;
}
 
// Driver Code
int main()
{
    int N = 5;
 
    // Given set of points with speed
    vector> D = { { 5, 12, 1 },
                              { 16, 63, 5 },
                              { -10, 24, 2 },
                              { 7, 24, 2 },
                              { -24, 7, 2 } };
 
    int X = 0, Y = 0;
 
    // Function call
    cout << (solve(D, N, X, Y));
 
    return 0;
}
 
// This code is contributed by mohit kumar 29

Java

// Java program to implement
// the above approach
import java.util.*;
import java.lang.*;
 
class GFG{
 
// Function to find the count of
// possible pairs of collisions
static double solve(int[][] D, int N,
                        int X, int Y)
{
     
    // Stores the time at which
    // points reach the origin
    ArrayList T = new ArrayList<>();
 
    // Calculate time for each point
    for(int i = 0; i < N; i++)
    {
        int x = D[i][0];
        int y = D[i][1];
 
        double speed = D[i][2];
 
        double time = ((x * x - X * X) +
                       (y * y - Y * Y)) /
                       (speed * speed);
 
        T.add(time);
    }
 
    // Sort the times
    Collections.sort(T);
 
    int i = 0;
    int total = 0;
 
    // Counting total collisions
    while (i < (T.size() - 1))
    {
         
        // Count of elements arriving at
        // a given point at the same time
        int count = 1;
 
        while ((i < (T.size() - 1)) &&
               (Double.compare(T.get(i),
                               T.get(i + 1)) == 0))
        {
            count += 1;
            i += 1;
        }
 
        total += (count * (count - 1)) / 2;
        i += 1;
         
    }
    return total;
}
 
// Driver Code
public static void main (String[] args)
{
    int arr[] = { 1, 2, 3, 4, 5 };
    int N = 5;
     
    // Given set of points with speed
    int[][] D = { { 5, 12, 1 },
                  { 16, 63, 5 },
                  { -10, 24, 2 },
                  { 7, 24, 2 },
                  { -24, 7, 2 } };
     
    int X = 0, Y = 0;
     
    // Function call
    System.out.println(solve(D, N, X, Y));
}
}
 
// This code is contributed by offbeat

Python3

# Python3 Program to implement
# the above approach
 
# Function to find the count of
# possible pairs of collisions
def solve(D, N, X, Y):
 
    # Stores the time at which
    # points reach the origin
    T = []
         
    # Calculate time for each point
    for i in range(N):
 
        x = D[i][0]
        y = D[i][1]
 
        speed = D[i][2]
 
        time = ((x * x - X * X) +
                (y * y - Y * Y)) / (speed * speed)
                 
 
        T.append(time)
         
    # Sort the times
    T.sort()
     
    i = 0
    total = 0
         
 
    # Counting total collisions
    while i

---

C#

// C# program to implement
// the above approach
using System;
using System.Collections;
 
class GFG{
 
// Function to find the count of
// possible pairs of collisions
static double solve(int[, ] D, int N,
                        int X, int Y)
{
     
    // Stores the time at which
    // points reach the origin
    ArrayList T = new ArrayList();
 
    // Calculate time for each point
    for(int i = 0; i < N; i++)
    {
        int x = D[i, 0];
        int y = D[i, 1];
 
        double speed = D[i, 2];
 
        double time = ((x * x - X * X) +
                       (y * y - Y * Y)) /
                       (speed * speed);
 
        T.Add(time);
    }
 
    // Sort the times
    T.Sort();
 
    int j = 0;
    int total = 0;
 
    // Counting total collisions
    while (j < (T.Count - 1))
    {
         
        // Count of elements arriving at
        // a given point at the same time
        int count = 1;
     
        while ((j < (T.Count - 1)) &&
              (Convert.ToDouble(T[j]).CompareTo(
               Convert.ToDouble(T[j + 1])) == 0))
        {
            count += 1;
            j += 1;
        }
        total += (count * (count - 1)) / 2;
        j += 1;
    }
    return total;
}
 
// Driver Code
public static void Main (String[] args)
{
    int N = 5;
     
    // Given set of points with speed
    int [,] D = new int [,] { { 5, 12, 1 },
                              { 16, 63, 5 },
                              { -10, 24, 2 },
                              { 7, 24, 2 },
                              { -24, 7, 2 } };
     
    int X = 0, Y = 0;
     
    // Function call
    Console.WriteLine(solve(D, N, X, Y));
}
}
 
// This code is contributed by jana_sayantan

Javascript

 
// JavaScript program to implement
// the above approach
 
// Function to find the count of
// possible pairs of collisions
function solve(D,N,X,Y)
{
    // Stores the time at which
    // points reach the origin
    let T = [];
   
    // Calculate time for each point
    for(let i = 0; i < N; i++)
    {
        let x = D[i][0];
        let y = D[i][1];
   
        let speed = D[i][2];
   
        let time = ((x * x - X * X) +
                       (y * y - Y * Y)) /
                       (speed * speed);
   
        T.push(time);
    }
   
    // Sort the times
    T.sort(function(a,b){return a-b;});
   
    let i = 0;
    let total = 0;
   
    // Counting total collisions
    while (i < (T.length - 1))
    {
           
        // Count of elements arriving at
        // a given point at the same time
        let count = 1;
   
        while ((i < (T.length - 1)) &&
               (T[i]==T[i+1]))
        {
            count += 1;
            i += 1;
        }
   
        total += (count * (count - 1)) / 2;
        i += 1;
           
    }
    return total;
}
 
// Driver Code
let arr=[1, 2, 3, 4, 5];
let N = 5;
 // Given set of points with speed
let D = [[5, 12, 1], [16, 63, 5],
    [-10, 24, 2], [7, 24, 2],
    [-24, 7, 2]];
 
let X = 0, Y = 0;
// Function call
document.write(solve(D, N, X, Y).toFixed(1));
 
// This code is contributed by patel2127
 

输出：

4.0

时间复杂度： O(N log N)
辅助空间： O(1)

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