查找覆盖给定数组中每个点的段数

给定线段和一些点，对于每个点，找到覆盖该点的线段数。

A segment (l, r) covers a point x if and only if l < = x < = r.

例子：

Input: Segments = {{0, 3}, {1, 3}, {3, 8}}, 
Points = {-1, 3, 8}.
Output : {0, 3, 1}
Explanation :
 

 

• No segments passing through point -1
• All the segments passing through point 3
• Segment 3rd passing through point 8

Input: Segments = {{1, 3}, {2, 4}, {5, 7}}, 
Points = {0, 2, 5}.
Output: {0, 2, 1}
Explanation :
 

 

• No segments passing through point 0
• 1st and 2nd segment passing through point 2
• Segment 3rd passing through point 5

方法：

• 我们可以通过使用类似于前缀和的逻辑来做到这一点。
• 让我们用(l，r)表示一个线段。形成一个成对的向量，对于每个段，在向量中推入两对，其值分别为(l，+1)和(r + 1，-1)。
• 按升序对点进行排序，但我们还需要它的位置，因此将其与它的位置映射在一起。
• 对段向量进行降序排序，因为我们从后面对其进行了迭代。
• 对段进行可变计数，该计数最初为零。
• 然后，我们将迭代该点，并从线段向量中弹出该对，直到其第一个值小于当前点，然后将其第二个值添加到计数中。
• 最后，将count的值存储在数组中的相应位置并打印该数组。

下面是上述方法的实现。

C++

// C++ program to find the number of
// segments covering each points
#include
using namespace std;
 
// Function to print an array
void PrintArray(int n,int arr[])
{
     for(int i = 0; i < n; i++)
     {
         cout< >segments,
                      vectorpoints, int s, int p)
{
   vector< pair >pts, seg;
     
   // Pushing points and index in
   // vector as a pairs
   for(int i = 0; i < p; i++)
   {
      pts.push_back({points[i], i});;
   }
     
   for(int i = 0; i < s; i++)
   {
      // (l,+1)
      seg.push_back({segments[i].first, 1});
      // (r+1,-1)
      seg.push_back({segments[i].second+1, -1});
   }
     
   // Sort the vectors
   sort(seg.begin(), seg.end(),
        greater>());
   sort(pts.begin(),pts.end());
     
   int count = 0;
   int ans[p];
     
   for(int i = 0; i < p; i++)
   {
        int x = pts[i].first;
        
        while(!seg.empty() &&
              seg.back().first <= x)
        {
            count+= seg.back().second;
            seg.pop_back();
        }
        ans[pts[i].second] = count;
   }
     
   // Print the answer
   PrintArray(p, ans);
   
}
 
//Driver code
int main()
{
  // Initializing vector of pairs
  vector>seg;
     
  // Push segments
  seg.push_back({0, 3});
  seg.push_back({1, 3});
  seg.push_back({3, 8});
     
  // Given points
  vectorpoint{-1, 3, 7};
     
  int s = seg.size();
  int p = point.size();
     
  NumberOfSegments(seg, point, s, p);
     
  return 0;
}

Java

// Java program to find the number of 
// segments covering each points
import java.util.*;
import java.lang.*;
 
class GFG{
  
// Function to print an array
static void PrintArray(int n,int arr[])
{
    for(int i = 0; i < n; i++)
    {
        System.out.print(arr[i] + " ");
    }
}
 
// Funtion prints number of segments
// covering by each points
static void NumberOfSegments(ArrayList segments,
                         int[] points, int s, int p)
{
    ArrayList pts = new ArrayList<>(),
                     seg = new ArrayList<>();
     
    // Pushing points and index in
    // vector as a pairs
    for(int i = 0; i < p; i++)
    {
        pts.add(new int[]{points[i], i});
    }
     
    for(int i = 0; i < s; i++)
    {
        // (l,+1)
        seg.add(new int[]{segments.get(i)[0], 1});
         
        // (r+1,-1)
        seg.add(new int[]{segments.get(i)[1] + 1, -1});
    }
     
    // Sort the vectors
    Collections.sort(seg, (a, b) -> b[0] - a[0]);
    Collections.sort(pts, (a, b) -> a[0] - b[0]);
     
    int count = 0;
    int[] ans = new int[p];
     
    for(int i = 0; i < p; i++)
    {
        int x = pts.get(i)[0];
         
        while (seg.size() != 0 &&
               seg.get(seg.size() - 1)[0] <= x)
        {
            count += seg.get(seg.size() - 1)[1];
            seg.remove(seg.size() - 1);
        }
        ans[pts.get(i)[1]] = count;
    }
     
    // Print the answer
    PrintArray(p, ans);
}
 
// Driver code
public static void main(String[] args)
{
     
    // Initializing vector of pairs
    ArrayListseg = new ArrayList<>();
     
    // Push segments
    seg.add(new int[]{0, 3});
    seg.add(new int[]{1, 3});
    seg.add(new int[]{3, 8});
     
    // Given points
    int[] point = {-1, 3, 7};
     
    int s = seg.size();
    int p = point.length;
     
    NumberOfSegments(seg, point, s, p);
}
}
 
// This code is contributed by offbeat

Python3

# Python3 program to find the number
# of segments covering each point
 
# Function to print an array
def PrintArray(n, arr):
     
    for i in range(n):
        print(arr[i], end = " ")
 
# Funtion prints number of segments
# covering by each points
def NumberOfSegments(segments, points, s, p):
     
    pts = []
    seg = []
     
    # Pushing points and index in
    # vector as a pairs
    for i in range(p):
        pts.append([points[i], i])
 
    for i in range(s):
         
        # (l, +1)
        seg.append([segments[i][0], 1])
         
        # (r+1, -1)
        seg.append([segments[i][1] + 1, -1])
 
    # Sort the vectors
    seg.sort(reverse = True)
    pts.sort(reverse = False)
     
    count = 0
    ans = [0 for i in range(p)]
 
    for i in range(p):
        x = pts[i][0]
 
        while(len(seg) != 0 and
          seg[len(seg) - 1][0] <= x):
                   
            count += seg[len(seg) - 1][1]
            seg.remove(seg[len(seg) - 1])
             
        ans[pts[i][1]] = count
         
    # Print the answer
    PrintArray(p, ans)
 
# Driver code
if __name__ == '__main__':
     
    # Initializing vector of pairs
    seg = []
 
    # Push segments
    seg.append([ 0, 3 ])
    seg.append([ 1, 3 ])
    seg.append([ 3, 8 ])
         
    # Given points
    point = [ -1, 3, 7 ]
         
    s = len(seg)
    p = len(point)
         
    NumberOfSegments(seg, point, s, p)
 
# This code is contributed by Bhupendra_Singh

输出：

0 3 1

时间复杂度：O(s + p)

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