[Ramda] Handle Errors in Ramda Pipelines with tryCatch

Handling your logic with composable functions makes your code declarative, leading to code that's easy to read and easy to test. Breaking that up to w
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POJ 1269 Intersecting Lines

直线求交,我的方法是叉积为0联立解方程。#include<iostream>#include<cstdio>#include<cstring>#include<algorithm>#include<cmath>#define maxn 10
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【SPFA+二分答案】BZOJ1614- [Usaco2007 Jan]Telephone Lines架设电话线

沉迷于刷水以前的那个二分写法过不了QAQ 换了一种好像大家都比较常用的二分。原因还不是很清楚。【题目大意】给出一张图,可以将其中k条边的边权减为0,求1到n的路径中最长边的最小值。【思路】二分答案,即最长边的最小值x。对于每次check(x),我们将边权大于x的边设为1,边权小于等于x的边设为0,跑
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Walkthrough: Arranging Controls on Windows Forms Using Snaplines

https://msdn.microsoft.com/en-us/library/t5b5kc41(v=vs.110).aspxSpacing and Aligning Controls Using SnaplinesSnaplines give you an accurate and intuit
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SmartSVN提示 svn: File has inconsistent newlines 解决

用SmartSVN提交代码的时候提示:svn: File has inconsistent newlines本文转自:http://www.youduoshao.com/2014-10-05/201410052335.html这是由于要提交的文件编码时混合了windows和unix符号导致的。解决方
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POJ 1269 - Intersecting Lines 直线与直线相交

题意:    判断直线间位置关系: 相交,平行,重合  1 include <iostream> 2 #include <cstdio> 3 using namespace std; 4 struct Point 5 { 6 i
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UVALive - 6955 Finding Lines 随机算法

题目链接:http://acm.hust.edu.cn/vjudge/contest/126968#problem/F题意给你n个点,问是否有>=p/100*n个点共线(p>=20&&p<=100)。题解随机枚举两个点,判过这两条直线的点是否满足要求。代码#incl
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POJ 1269 Intersecting Lines【判断直线相交】

题意:给两条直线,判断相交,重合或者平行思路:判断重合可以用叉积,平行用斜率,其他情况即为相交。求交点:这里也用到叉积的原理。假设交点为p0(x0,y0)。则有:(p1-p0)X(p2-p0)=0(p3-p0)X(p2-p0)=0展开后即是(y1-y2)x0+(x2-x1)y0+x1y2-x2y1=
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outdated: 21.Lines, Antialiasing, Timing, Ortho View And Simple Sounds

这是一个简单的小游戏,类型相当于amidar,不过也有1000行,手困眼花......UP\DOWN\LEFT\RIGHT控制,SPACE重新开始。在TimerInit()函数中,QueryPerformanceFrequency()函数查询性能计数器的频率。QueryPerformanceCoun
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BZOJ_1614_ [Usaco2007_Jan]_Telephone_Lines_架设电话线_(二分+最短路_Dijkstra/Spfa)

描述http://www.lydsy.com/JudgeOnline/problem.php?id=1614 分析类似POJ_3662_Telephone_Lines_(二分+最短路) Dijkstra: 1 #include <bits/stdc++.h> 2 us
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