给定 2D 平面上点的n 个坐标 (x, y) 和 Q 查询。每个查询包含一个整数r ，任务是计算位于半径为 r 且以原点为中心的圆的内部或圆周上的点数。
例子 ：
Input : n = 5
Coordinates:
1 1
2 2
3 3
-1 -1
4 4
Query 1: 3
Query 2: 32
Output :
3
5
For first query radius = 3, number of points lie
inside or on the circumference are (1, 1), (-1, -1),
(2, 2). There are only 3 points lie inside or on
the circumference of the circle.
For second query radius = 32, all five points are
inside the circle. 以原点 (0, 0) 为圆心，半径为 r, x 2 + y 2 = r 2的圆的方程。并且条件 (x 1 , y 1 ) 位于圆周内或圆周上，x 1 2 + y 1 2 <= r 2 。
Naive 方法可以针对每个查询，遍历所有点并检查条件。这需要 O(n*Q) 时间复杂度。
一种有效的方法是为每个点坐标预先计算 x 2 + y 2并将它们存储在数组 p[] 中。现在，对数组 p[] 进行排序。然后对数组应用二分搜索以找到每个查询的条件 p[i] <= r 2 的最后一个索引。
下面是这个方法的实现：
C++
// C++ program to find number of points lie inside or
// on the cirumference of circle for Q queries.
#include
using namespace std;
// Computing the x^2 + y^2 for each given points
// and sorting them.
void preprocess(int p[], int x[], int y[], int n)
{
for (int i = 0; i < n; i++)
p[i] = x[i] * x[i] + y[i] * y[i];
sort(p, p + n);
}
// Return count of points lie inside or on circumference
// of circle using binary search on p[0..n-1]
int query(int p[], int n, int rad)
{
int start = 0, end = n - 1;
while ((end - start) > 1) {
int mid = (start + end) / 2;
double tp = sqrt(p[mid]);
if (tp > (rad * 1.0))
end = mid - 1;
else
start = mid;
}
double tp1 = sqrt(p[start]), tp2 = sqrt(p[end]);
if (tp1 > (rad * 1.0))
return 0;
else if (tp2 <= (rad * 1.0))
return end + 1;
else
return start + 1;
}
// Driven Program
int main()
{
int x[] = { 1, 2, 3, -1, 4 };
int y[] = { 1, 2, 3, -1, 4 };
int n = sizeof(x) / sizeof(x[0]);
// Compute distances of all points and keep
// the distances sorted so that query can
// work in O(logn) using Binary Search.
int p[n];
preprocess(p, x, y, n);
// Print number of points in a circle of radius 3.
cout << query(p, n, 3) << endl;
// Print number of points in a circle of radius 32.
cout << query(p, n, 32) << endl;
return 0;
} Java
// JAVA Code for Queries on count of
// points lie inside a circle
import java.util.*;
class GFG {
// Computing the x^2 + y^2 for each given points
// and sorting them.
public static void preprocess(int p[], int x[],
int y[], int n)
{
for (int i = 0; i < n; i++)
p[i] = x[i] * x[i] + y[i] * y[i];
Arrays.sort(p);
}
// Return count of points lie inside or on
// circumference of circle using binary
// search on p[0..n-1]
public static int query(int p[], int n, int rad)
{
int start = 0, end = n - 1;
while ((end - start) > 1) {
int mid = (start + end) / 2;
double tp = Math.sqrt(p[mid]);
if (tp > (rad * 1.0))
end = mid - 1;
else
start = mid;
}
double tp1 = Math.sqrt(p[start]);
double tp2 = Math.sqrt(p[end]);
if (tp1 > (rad * 1.0))
return 0;
else if (tp2 <= (rad * 1.0))
return end + 1;
else
return start + 1;
}
/* Driver program to test above function */
public static void main(String[] args)
{
int x[] = { 1, 2, 3, -1, 4 };
int y[] = { 1, 2, 3, -1, 4 };
int n = x.length;
// Compute distances of all points and keep
// the distances sorted so that query can
// work in O(logn) using Binary Search.
int p[] = new int[n];
preprocess(p, x, y, n);
// Print number of points in a circle of
// radius 3.
System.out.println(query(p, n, 3));
// Print number of points in a circle of
// radius 32.
System.out.println(query(p, n, 32));
}
}
// This code is contributed by Arnav Kr. Mandal.Python 3
# Python 3 program to find number of
# points lie inside or on the cirumference
# of circle for Q queries.
import math
# Computing the x^2 + y^2 for each
# given points and sorting them.
def preprocess(p, x, y, n):
for i in range(n):
p[i] = x[i] * x[i] + y[i] * y[i]
p.sort()
# Return count of points lie inside
# or on circumference of circle using
# binary search on p[0..n-1]
def query(p, n, rad):
start = 0
end = n - 1
while ((end - start) > 1):
mid = (start + end) // 2
tp = math.sqrt(p[mid])
if (tp > (rad * 1.0)):
end = mid - 1
else:
start = mid
tp1 = math.sqrt(p[start])
tp2 = math.sqrt(p[end])
if (tp1 > (rad * 1.0)):
return 0
elif (tp2 <= (rad * 1.0)):
return end + 1
else:
return start + 1
# Driver Code
if __name__ == "__main__":
x = [ 1, 2, 3, -1, 4 ]
y = [ 1, 2, 3, -1, 4 ]
n = len(x)
# Compute distances of all points and keep
# the distances sorted so that query can
# work in O(logn) using Binary Search.
p = [0] * n
preprocess(p, x, y, n)
# Print number of points in a
# circle of radius 3.
print(query(p, n, 3))
# Print number of points in a
# circle of radius 32.
print(query(p, n, 32))
# This code is contributed by ita_cC#
// C# Code for Queries on count of
// points lie inside a circle
using System;
class GFG {
// Computing the x^2 + y^2 for each
// given points and sorting them.
public static void preprocess(int[] p, int[] x,
int[] y, int n)
{
for (int i = 0; i < n; i++)
p[i] = x[i] * x[i] + y[i] * y[i];
Array.Sort(p);
}
// Return count of points lie inside or on
// circumference of circle using binary
// search on p[0..n-1]
public static int query(int[] p, int n, int rad)
{
int start = 0, end = n - 1;
while ((end - start) > 1) {
int mid = (start + end) / 2;
double tp = Math.Sqrt(p[mid]);
if (tp > (rad * 1.0))
end = mid - 1;
else
start = mid;
}
double tp1 = Math.Sqrt(p[start]);
double tp2 = Math.Sqrt(p[end]);
if (tp1 > (rad * 1.0))
return 0;
else if (tp2 <= (rad * 1.0))
return end + 1;
else
return start + 1;
}
/* Driver program to test above function */
public static void Main()
{
int[] x = { 1, 2, 3, -1, 4 };
int[] y = { 1, 2, 3, -1, 4 };
int n = x.Length;
// Compute distances of all points and keep
// the distances sorted so that query can
// work in O(logn) using Binary Search.
int[] p = new int[n];
preprocess(p, x, y, n);
// Print number of points in a circle of
// radius 3.
Console.WriteLine(query(p, n, 3));
// Print number of points in a circle of
// radius 32.
Console.WriteLine(query(p, n, 32));
}
}
// This code is contributed by vt_m.Javascript
输出：
3
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