代码编写

这次的作业瞬间难了好多,无论是问题本身的难度或者是单元测试这一原来没接触过的概念或者是命令行参数的处理这些琐碎的问题,都使得这次作业的完成说不上轻松。

最大子数组之和垂直水平相连的拓展问题解决关键在于循环语句的适度改写,连通问题则是用递归搜索的方法来解决(效率没有实测),在15*15的情况下还是能较快得出结果的。

非常庆幸使用的是Python,Pyhton中很多语法能够保证我在编写代码时不用分太多的时间去处理数据输入,在处理问题上一些数组相关灵活的语法也很大程度上方便了代码的编写。

 # coding:utf-8
'''
2013-9-30 XTH
'''
import sys def setglobalvar():
global max_sum,now_sum,min_x,min_y,num,visited,pointgroup
max_sum = 0
now_sum = 0
min_x = 0
min_y = 0
num = []
visited = {}
pointgroup = [] def maxsum_h(num,n1,n2):#水平上相连
line = [0]*n2
max_sum = 0 #最大和
now_sum = 0 #当前和
for l in range (0,n2):
for i in range(0,n1):
for j in range(i,n1):
for k in range(0+l,n2+l):
k = k % n2
line[k] += num[j][k]
if now_sum <0:
now_sum = 0
now_sum += line[k]
if now_sum > max_sum:
max_sum = now_sum
now_sum = 0
now_sum=0
line = [0]*n2
return max_sum def maxsum_v(num,n1,n2): #垂直上相连
line = [0]*n2
max_sum = 0 #最大和
now_sum = 0 #当前和
for l in range (0,n1):
for i in range(0,n1):
for j in range(i+l,n1+l):
for k in range(0,n2):
j = j % n1
line[k] += num[j][k]
if now_sum <0:
now_sum = 0
now_sum += line[k]
if now_sum > max_sum:
max_sum = now_sum
now_sum = 0
now_sum=0
line = [0]*n2
return max_sum def maxsum(num,n1,n2):#普通
line = [0]*n2
max_sum = 0 #最大和
now_sum = 0 #当前和
for i in range(0,n1):
for j in range(i,n1):
for k in range(0,n2):
line[k] += num[j][k]
if now_sum <0:
now_sum = 0
now_sum += line[k]
if now_sum > max_sum:
max_sum = now_sum
now_sum = 0
now_sum=0
line = [0]*n2
return max_sum def maxsum_vh(num,n1,n2):#垂直水平相连
line = [0]*n2
max_sum = 0 #最大和
now_sum = 0 #当前和
for l1 in range (0,n1):
for l2 in range (0,n2):
for i in range(0,n1):
for j in range(i+l1,n1+l1):
for k in range(0+l2,n2+l2):
j = j % n1
k = k % n2
line[k] += num[j][k]
if now_sum <0:
now_sum = 0
now_sum += line[k]
if now_sum > max_sum:
max_sum = now_sum
now_sum = 0
now_sum=0
line = [0]*n2
return max_sum def searchthrough(x,y,num,now_sum):#搜索函数
global max_sum,pointgroup,min_x,min_y,visited
max_sum = max(max_sum, now_sum)
for i in [[0,-1],[1,0],[0,1],[-1,0]]:
if x+i[0]>=min_x and x+i[0]<n1 and y+i[1]>=min_y and y+i[1]<n2 and visited[(x+i[0])%n1,(y+i[1])%n2]==0 and [(x+i[0])%n1,(y+i[1])%n2,num[(x+i[0])% n1][(y+i[1])%n2]] not in pointgroup:
pointgroup.append([(x + i[0]) % n1, (y + i[1]) % n2, num[(x + i[0]) % n1][(y + i[1]) % n2]])
if pointgroup == []:
return
pointgroup = sorted(pointgroup, key=lambda x: x[2])
nextpoint = pointgroup.pop()
if now_sum + nextpoint[2] > 0:
visited[nextpoint[0], nextpoint[1]] = 1
searchthrough(nextpoint[0],nextpoint[1],num,now_sum + nextpoint[2])
visited[nextpoint[0], nextpoint[1]] = 0
else:
return def maxsum_a(num,n1,n2): #连通
global min_x,min_y,max_sum,visited
min_x = 0
min_y = 0
max_sum = 0
now_sum = 0
startpointx = []
startpointy = []
pointgroup = []
for i in range(0,n1):
for j in range(0,n2):
visited[i,j] = 0
for i in range(0,n1):
for j in range(0,n2):
if num[i][j] > 0:
startpointx.append(i)
startpointy.append(j)
for pointx in startpointx:
pointy = startpointy.pop()
visited[pointx, pointy] = 1
searchthrough(pointx,pointy,num,num[pointx][pointy])
return max_sum def maxsum_vha(num,n1,n2): #水平垂直上相连 连通
global min_x,min_y,max_sum,visited
min_x = -n1
min_y = -n2
max_sum = 0
now_sum = 0
startpointx = []
startpointy = []
pointgroup = []
for i in range(0,n1):
for j in range(0,n2):
visited[i,j] = 0
for i in range(0,n1):
for j in range(0,n2):
if num[i][j] > 0:
startpointx.append(i)
startpointy.append(j)
for pointx in startpointx:
pointy = startpointy.pop()
visited[pointx, pointy] = 1
searchthrough(pointx,pointy,num,num[pointx][pointy])
return max_sum def main():
setglobalvar()
global n1,n2
max_sum = 0
V = H = A = False
if "\\v" in sys.argv[1:]:
V = True;
if "\\h" in sys.argv[1:]:
H = True;
if "\\a" in sys.argv[1:]:
A = True;
filename = sys.argv[-1];
try:
f = open(filename,"r")
except:
raise IOError("ERROR:can't open the file")
try:
line = f.readline()
line = line.strip('\n').strip(',')
n1 = int(line)
line = f.readline()
line = line.strip('\n').strip(',')
n2 = int(line)
num=[[]]*int(n1)
for i in range(0,int(n1)):
line = f.readline()
line = line.strip('\n')
if len(line.split(",")) != n2:
raise ValueError("ERROR:the format of file is wrong")
num[i] = line.split(",")
num=[[int(x) for x in inner] for inner in num]
except:
raise ValueError("ERROR:the format of file is wrong")
if V!=True and H!=True and A == True:#连通
max_sum = maxsum_a(num,n1,n2);
elif V==True and H!=True and A != True:#水平上相连
max_sum = maxsum_v(num,n1,n2);
elif V!=True and H==True and A != True:#垂直上相连
max_sum = maxsum_h(num,n1,n2);
elif V==True and H==True and A != True:#水平垂直上相连
max_sum = maxsum_vh(num,n1,n2);
elif V==True and H==True and A == True:#水平垂直上相连连通
max_sum = maxsum_vha(num,n1,n2);
else:#普通
max_sum = maxsum(num,n1,n2);
return max_sum if __name__ == '__main__':
print main()

单元测试

我大概明白单元测试的概念,但是本次问题很难被看做是一个模块,写单元测试的时候也无从下手,只是简单地测试了一下命令行参数的处理、以及样例的结果验证。

项目时间

PSP2.1

Personal Software Process Stages

Time (%) Senior Student

Planning

计划

6

·         Estimate

·         估计这个任务需要多少时间

6

Development

开发

80

·         Analysis

·         需求分析 (包括学习新技术)

10

·         Design Spec

·         生成设计文档

0

·         Design Review

·         设计复审 (和同事审核设计文档)

0

·         Coding Standard

·         代码规范 (为目前的开发制定合适的规范)

0

·         Design

·         具体设计

15

·         Coding

·         具体编码

40

·         Code Review

·         代码复审

5

·         Test

·         测试(自我测试,修改代码,提交修改)

10

Reporting

报告

14

  • ·         Test Report
  • ·         测试报告
 

2

  • ·         Size Measurement
  • ·         计算工作量
 

2

  • ·         Postmortem & Process Improvement Plan
  • ·         事后总结, 并提出过程改进计划
 

10

总结

由于进入大学之后没有搞过ACM,在解决这个问题的时候非常得吃力,对于群里大家提出的想法也只能是不明觉厉,希望能够在接下来的课程中多多提高吧。

homework-02 "最大子数组之和"的问题进阶的更多相关文章

  1. homework-01 "最大子数组之和"的问题求解过程

    写在前面:我的算法能力很弱,并且也是第一次写博文,总之希望自己能在这次的课程中学到很多贴近实践的东西吧. 1.这次的程序是python写的,这也算是我第一次正正经经地拿python来写东西,结果上来说 ...

  2. 求一个二维整数数组最大子数组之和,时间复杂度为N^2

    本随笔只由于时间原因,我就只写写思想了 二维数组最大子数组之和,可以  引用  一维最大子数组之和 的思想一维最大子数组之和 的思想,在本博客上有,这里就不做多的介绍了 我们有一个最初的二维数组a[n ...

  3. [LeetCode] Maximum Size Subarray Sum Equals k 最大子数组之和为k

    Given an array nums and a target value k, find the maximum length of a subarray that sums to k. If t ...

  4. 求一个整数数组最大子数组之和,时间复杂度为N

    #include<iostream.h> int main () { ]={-,-,-,-,-,-,-,-,-,-}; ],sum=; ;i<;i++) { ) { sum=b[i] ...

  5. homework-01 "最大子数组之和"的解决过程

    看到这个题目,我首先想到就是暴力解决 求出所有的子数组的和,取出最大值即可 但其中是可以有优化的 如 子数组[3:6]可以用[3:5]+[6]来计算 即可以将前面的计算结果保留下来,减少后面的重复计算 ...

  6. [Swift]LeetCode325. 最大子数组之和为k $ Maximum Size Subarray Sum Equals k

    Given an array nums and a target value k, find the maximum length of a subarray that sums to k. If t ...

  7. 最大子数组之和(N)

    int maxSum(int *array, int n) { ]; ; ; ; i < n; i++) { ) newsum += array[i]; else newsum = array[ ...

  8. [LeetCode] Maximum Product Subarray 求最大子数组乘积

    Find the contiguous subarray within an array (containing at least one number) which has the largest ...

  9. [LeetCode] Maximum Subarray 最大子数组

    Find the contiguous subarray within an array (containing at least one number) which has the largest ...

随机推荐

  1. [Mongo] error inserting documents: BSONObj size is invalid (mongoimport mongorestore 数据备份恢复)

    解决办法如下, ./mongoimport -port 6066 -d xxx -c xxx --batchSize=10 /root/mong_data/test/xxx 原因转自 http://b ...

  2. NuGet在2015中的使用

    NuGet Package Restore  https://docs.nuget.org/Consume/Package-Restore 以https://github.com/andburn/hd ...

  3. tranform-scale 缩小元素,移上去文字抖动

    元素缩小后,鼠标移上去之后文字会出现抖动, -webkit-transform:scale(0.5); 修复代码如下: *{ -webkit-backface-visibility: hidden; ...

  4. java 语言里 遍历 collection 的方式

    我来简单说一下吧,一般有2种方法来遍历collection中的元素,以HashSet为例子HashSet hs=new HashSet();hs.add("hello");hs.a ...

  5. CodeForces Round #278 (Div.2) (待续)

    A 这么简单的题直接贴代码好了. #include <cstdio> #include <cmath> using namespace std; bool islucky(in ...

  6. bzoj1927: [Sdoi2010]星际竞速

    跟上一题几乎一样... #include<cstdio> #include<cstring> #include<iostream> #include<algo ...

  7. 转:ASP.NET MVC中Unobtrusive Ajax的妙用

    Unobtrusive Javascript有三层含义:一是在HTML代码中不会随意的插入Javsscript代码,只在标签中加一些额外的属性值,然后被引用的脚本文件识别和处理:二是通过脚本文件所增加 ...

  8. dbms_stats.gather_table_stats与analyze table 的区别[转贴]

    Analyze StatementThe ANALYZE statement can be used to gather statistics for a specific table, index ...

  9. memcache分布式部署的原理分析

    下面本文章来给各位同学介绍memcache分布式部署的原理分析,希望此文章对你理解memcache分布式部署会有所帮助哦.   今天在封装memcache操作类库过程中,意识到一直以来对memcach ...

  10. android模块化app开发笔记-2插件间布局文件共享

    android编程时布局文件,图片资源等都是放在同一个文件夹下,这样照成一个问题就是我们想重用UI布局文件和图片时就还需要其分离这些资料,相信大部分android程序员都遇到过这样的问题,其痛苦程度不 ...