Problem Introduction

The goal in this problem is given a set of segments on a line and a set of points on a line, to count, for each point, the number of segments which contain it.

Problem Description

Task.In this problem you are given a set of points on a line and a set of segments on a line. The goal is to compute, for each point, the number of segments that contain this point.

Input Format.The first line contains two non-negative integers \(s\) and \(p\) defining the number of segments and the number of points on a line, respectively. The next \(s\) lines contain two integers \(a_i, b_i\) defining the \(i\)-th segment \([a_i, b_i]\). The next line contains \(p\) integers defining points \(x_1,x_2,\cdots ,x_p\).

Constraints.\(1 \leq s,p \leq 50000;-10^8 \leq a_i \leq b_i \leq 10^8\) for all \(0 \leq i < s;-10^8 \leq x_j \leq 10^8\) for all \(0 \leq j < p\).

Output Format.Output \(p\) non-negative integers \(k_0,k_1,\cdots,k_{p-1}\) where \(k_i\) is the number of segments which contain \(x_i\). More formally,
\(k_i=|{j:a_j \leq x_i \leq b_j}|\)

Sample 1.
Input:

2 3
0 5
7 10
1 6 11

Output:

1 0 0

Sample 2.
Input:

1 3
-10 10
-100 100 0

Output:

0 0 1

Sample 3.
Input:

3 2
0 5
-3 2
7 10
1 6

Output:

2 0

Solution

# Uses python3
import sys
from itertools import chain

def fast_count_segments(starts, ends, points):
    cnt = [0] * len(points)
    a = zip(starts, [float('-inf')]*len(starts))
    b = zip(ends, [float('inf')]*len(ends))
    c = zip(points, range(len(points)))
    sortedlist = sorted(chain(a,b,c), key=lambda a : (a[0], a[1]))
    stack = []
    for i, j in sortedlist:
        if j == float('-inf'):
            stack.append(j)
        elif j == float('inf'):
            stack.pop()
        else:
            cnt[j] = len(stack)
    return cnt

def naive_count_segments(starts, ends, points):
    cnt = [0] * len(points)
    for i in range(len(points)):
        for j in range(len(starts)):
            if starts[j] <= points[i] <= ends[j]:
                cnt[i] += 1
    return cnt

if __name__ == '__main__':
    input = sys.stdin.read()
    data = list(map(int, input.split()))
    n = data[0]
    m = data[1]
    starts = data[2:2 * n + 2:2]
    ends   = data[3:2 * n + 2:2]
    points = data[2 * n + 2:]
    #use fast_count_segments
    cnt = fast_count_segments(starts, ends, points)
    for x in cnt:
        print(x, end=' ')

[UCSD白板题] Points and Segments的更多相关文章

  1. [UCSD白板题] Covering Segments by Points

    Problem Introduction You are given a set of segments on a line and your goal is to mark as few point ...

  2. [UCSD白板题] Longest Common Subsequence of Three Sequences

    Problem Introduction In this problem, your goal is to compute the length of a longest common subsequ ...

  3. [UCSD白板题] Maximize the Value of an Arithmetic Expression

    Problem Introduction In the problem, your goal is to add parentheses to a given arithmetic expressio ...

  4. [UCSD白板题] Compute the Edit Distance Between Two Strings

    Problem Introduction The edit distinct between two strings is the minimum number of insertions, dele ...

  5. [UCSD白板题] Take as Much Gold as Possible

    Problem Introduction This problem is about implementing an algorithm for the knapsack without repeti ...

  6. [UCSD白板题] Primitive Calculator

    Problem Introduction You are given a primitive calculator that can perform the following three opera ...

  7. [UCSD白板题] Number of Inversions

    Problem Introduction An inversion of a sequence \(a_0,a_1,\cdots,a_{n-1}\) is a pair of indices \(0 ...

  8. [UCSD白板题] Sorting: 3-Way Partition

    Problem Introduction The goal in this problem is to redesign a given implementation of the randomize ...

  9. [UCSD白板题] Majority Element

    Problem Introduction An element of a sequence of length \(n\) is called a majority element if it app ...

随机推荐

  1. CentOS 7 安装 MySQL

    转自:http://www.centoscn.com/mysql/2016/0315/6844.html 环境 CentOS 7.1 (64-bit system) MySQL 5.6.24 Cent ...

  2. Js 职责链模式 简单理解

    js 职责链模式 的简单理解.大叔的代码太高深了,不好理解. function Handler(s) { this.successor = s || null; this.handle = funct ...

  3. sql遍历

    DECLARE @BTime DATETIME,@ETime DATETIME;DECLARE @Temp TABLE (ID BIGINT IDENTITY(1,1),aid BIGINT,newc ...

  4. 简单的html和css

    整体图太大了,看不太清楚,下面是分开的图 第一张: 第二张:

  5. myeclipse 手动安装 lombok

    1. 将 lombok.jar 复制到 myeclipse.ini / eclipse.ini 所在的文件夹目录下    2. 打开 eclipse.ini / myeclipse.ini,在最后面插 ...

  6. JSP中的 HttpSession、pageContext对象

    pageContext 隐含对象对应javax.servlet.jsp.PageContext,都自动的被加入至pageContext中, 您可以由它来取得与JSP相关的对应之Servlet对象,像是 ...

  7. Xml与DataTable相互转换方法

    1.Xml与DataTable相互转换方法:http://www.cnblogs.com/lilin/archive/2010/04/18/1714927.html

  8. UDP异步通信

    先看效果图 Server: using System; using System.Collections.Generic; using System.Text; using System.Net; u ...

  9. WebMethod在webservice里面非静态方法能调用,在页面类里面,静态方法才能调用

    WebMethod在webservice里面非静态方法能调用,在页面类里面,静态方法才能调用

  10. VLC嵌入网页,终于要成功了!

    <OBJECT classid="clsid:9BE31822-FDAD-461B-AD51-BE1D1C159921" width="640" heig ...