最大化相交线段的数量

给定两个数组X[]和Y[] ，代表X和Y数轴上的点，使得每个相似索引的数组元素形成一个线段，即X[i]和Y[i]形成一个线段，任务是找到可以从给定数组中选择的最大线段数。

例子：

Input: X[] = {0, 3, 4, 1, 2}, Y[] = {3, 2, 0, 1, 4}
Output: 3
Explanation: The set of line segments are {{1, 1}, {4, 0}, {0, 3}}.

Input: X[] = {1, 2, 0}, Y[] = {2, 0, 1}
Output: 2
Explanation: The set of line segments is {{1, 2}, {2, 0}}.

编程需要懂一点英语

方法：这个问题可以通过观察两个线段 (i, j) 之间的交集来解决，只有当X[i] < X[j]和Y[i] > Y[j] 时，反之亦然。因此，可以使用Sorting with Binary search解决问题，使用Sets可以找到这样的线段。

请按照以下步骤解决给定的问题：

初始化一个成对向量，比如p来存储对{X[i], Y[i]}作为一个元素。
按 X 数轴上点的升序对p对的向量进行排序，因此每条线段i 都满足交点的第一个条件，即X[k] < X[i]其中k < i 。
初始化一个 Set，比如s ，以降序存储Y[i]的值。
从p的第一个元素，将 Y 坐标（即p[0].second ）推入集合中。
迭代p 的所有元素，并针对每个元素：
- 执行二分搜索以找到p[i].second 的下界。
- 如果没有获得下界，则意味着p[i].second小于 Set 中存在的所有元素。这满足Y[i] < Y[k]其中k < i的第二个条件，因此将p[i].second推入Set 。
- 如果找到下界，则将其删除并将p[i].second推入 Set 中。
最后，返回作为结果的集合的大小。

下面是上述方法的实现：

C++

// C++ program for the above approach
#include 
using namespace std;
 
// Function to find the maximum number of
// intersecting line segments possible
int solve(int N, int X[], int Y[])
{
    // Stores pairs of line
    // segments {X[i], Y[i])
    vector > p;
 
    for (int i = 0; i < N; i++) {
        // Push {X[i], Y[i]} into p
        p.push_back({ X[i], Y[i] });
    }
 
    // Sort p in ascending order
    // of points on X number line
    sort(p.begin(), p.end());
 
    // Stores the points on Y number
    // line in descending order
    set > s;
 
    // Insert the first Y point from p
    s.insert(p[0].second);
 
    for (int i = 0; i < N; i++) {
 
        // Binary search to find the
        // lower bound of p[i].second
        auto it = s.lower_bound(p[i].second);
 
        // If lower_bound doesn't exist
        if (it == s.end()) {
 
            // Insert p[i].second into the set
            s.insert(p[i].second);
        }
        else {
 
            // Erase the next lower
            //_bound from the set
            s.erase(*it);
 
            // Insert p[i].second
            // into the set
            s.insert(p[i].second);
        }
    }
 
    // Return the size of the set
    // as the final result
    return s.size();
}
 
// Driver Code
int main()
{
    // Given Input
    int N = 3;
    int X[] = { 1, 2, 0 };
    int Y[] = { 2, 0, 1 };
 
    // Function call to find the maximum
    // number of intersecting line segments
    int maxintersection = solve(N, X, Y);
    cout << maxintersection;
}

Java

// Java program for the above approach
import java.io.*;
import java.lang.*;
import java.util.*;
 
class GFG{
 
// Function to find the maximum number of
// intersecting line segments possible
static int solve(int N, int X[], int Y[])
{
     
    // Stores pairs of line
    // segments {X[i], Y[i])
    ArrayList p = new ArrayList<>();
 
    for(int i = 0; i < N; i++)
    {
         
        // Add {X[i], Y[i]} into p
        p.add(new int[] { X[i], Y[i] });
    }
 
    // Sort p in ascending order
    // of points on X number line
    Collections.sort(p, (p1, p2) -> {
        if (p1[0] != p2[0])
            return p1[0] - p2[0];
             
        return p1[1] - p2[1];
    });
 
    // Stores the points on Y number
    // line in ascending order
    TreeSet s = new TreeSet<>();
 
    // Insert the first Y point from p
    s.add(p.get(0)[1]);
 
    for(int i = 0; i < N; i++)
    {
         
        // Binary search to find the
        // floor value of p.get(i)[1]
        Integer it = s.floor(p.get(i)[1]);
 
        // If floor value doesn't exist
        if (it == null)
        {
 
            // Insert p.get(i)[1] into the set
            s.add(p.get(i)[1]);
        }
        else
        {
             
            // Erase the next floor
            // value from the set
            s.remove(it);
 
            // Insert p.get(i)[1]
            // into the set
            s.add(p.get(i)[1]);
        }
    }
 
    // Return the size of the set
    // as the final result
    return s.size();
}
 
// Driver Code
public static void main(String[] args)
{
     
    // Given Input
    int N = 3;
    int X[] = { 1, 2, 0 };
    int Y[] = { 2, 0, 1 };
 
    // Function call to find the maximum
    // number of intersecting line segments
    int maxintersection = solve(N, X, Y);
    System.out.println(maxintersection);
}
}
 
// This code is contributed by Kingash

Python3

# Python3 program for the above approach
from bisect import bisect_left
 
# Function to find the maximum number of
# intersecting line segments possible
def solve(N, X, Y):
     
    # Stores pairs of line
    # segments {X[i], Y[i])
    p = []
 
    for i in range(N):
         
        # Push {X[i], Y[i]} into p
        p.append([X[i], Y[i]])
 
    # Sort p in ascending order
    # of points on X number line
    p = sorted(p)
 
    # Stores the points on Y number
    # line in descending order
    s = {}
 
    # Insert the first Y pofrom p
    s[p[0][1]] = 1
 
    for i in range(N):
         
        # Binary search to find the
        # lower bound of p[i][1]
        arr = list(s.keys())
 
        it = bisect_left(arr, p[i][1])
 
        # If lower_bound doesn't exist
        if (it == len(s)):
             
            # Insert p[i][1] into the set
            s[p[i][1]] = 1
        else:
 
            # Erase the next lower
            # _bound from the set
            del s[arr[it]]
 
            # Insert p[i][1]
            # into the set
            s[p[i][1]] = 1
 
    # Return the size of the set
    # as the final result
    return len(s)
 
# Driver Code
if __name__ == '__main__':
     
    # Given Input
    N = 3
    X = [1, 2, 0]
    Y = [2, 0, 1]
 
    # Function call to find the maximum
    # number of intersecting line segments
    maxintersection = solve(N, X, Y)
     
    print (maxintersection)
 
# This code is contributed by mohit kumar 29

输出：

时间复杂度： O(N log N)
辅助空间： O(N)

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