[UCSD白板题] Points and Segments
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
- [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 ...
- [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 ...
- [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 ...
- [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 ...
- [UCSD白板题] Take as Much Gold as Possible
Problem Introduction This problem is about implementing an algorithm for the knapsack without repeti ...
- [UCSD白板题] Primitive Calculator
Problem Introduction You are given a primitive calculator that can perform the following three opera ...
- [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 ...
- [UCSD白板题] Sorting: 3-Way Partition
Problem Introduction The goal in this problem is to redesign a given implementation of the randomize ...
- [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
- CC1310电源管脚
对于48pin脚的CC1310而言,属于电源类的管脚如下: 上述电源类管脚的关系如下: 1 VDDS类管脚 VDDS类管脚包括VDDS.VDDS2.VDDS3和VDDS_DCDC四个管脚.其中VDDS ...
- 安装android
http://www.oschina.net/question/1463998_220998 http://www.cnblogs.com/zoupeiyang/p/4034517.html
- notepad++ 离线插件下载
http://www.cnblogs.com/findumars/p/5180562.html
- lamp php的ssl,ssh支持
Php支持ssl,ssh扩展: 准备:可以成功解析php 1.curl的安装 [root@localhost~]# cd /usr/local/src/ [root@localhost~]# wget ...
- KVM 虚拟化 初体验
KVM 是 Kernel-based Virtual Machine 的简称,是 Linux 下 x86 硬件平台上的全功能虚拟化解决方案: 使用 KVM ,可允许运行多个虚拟机,包括 Linux 和 ...
- bootstrap的html模版
<!DOCTYPE html> <html> <head> <title>Bootstrap 模板</title> <meta nam ...
- Nginx密码验证 ngx_http_auth_basic_module模块
有时候我们需要限制某些目录只允许指定的用户才可以访问,我们可以给指定的目录添加一个用户限制. nginx给我们提供了ngx_http_auth_basic_module模块来实现这个功能. 模块ngx ...
- C# 将sheet中数据转为list
public IList<T> ExportToList<T>(ISheet sheet, string[] fields) where T : class,new() { I ...
- 《Linux 多线程服务端编程:使用 muduo C++ 网络库》电子版上市
<Linux 多线程服务端编程:使用 muduo C++ 网络库> 电子版已在京东和亚马逊上市销售. 京东购买地址:http://e.jd.com/30149978.html 亚马逊Kin ...
- LINUX 如何开放端口和关闭端口/jps/sudo命令
1 在java的根目录下用java的jps查看:============================================================================ ...