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【二分图匹配入门专题1】D - Matrix hdu2119【最小顶点覆盖】
Give you a matrix(only contains 0 or 1),every time you can select a row or a column and delete all the '1' in this row or this column . Your task is t ...
分类:其他好文   时间:2017-08-13 10:08:15    阅读次数:139
Logistic回归python实现
2017-08-12 Logistic 回归,作为分类器: 分别用了梯度上升,牛顿法来最优化损失函数: # -*- coding: utf-8 -*-'''function: 实现Logistic回归,拟合直线,对数据进行分类;利用梯度上升,随机梯度上升,改进的随机梯度上升,牛顿法分别对损失函数优化 ...
分类:编程语言   时间:2017-08-13 00:18:32    阅读次数:347
试着用React写项目-利用react-router解决跳转路由等问题(二)
转载请注明出处:王亟亟的大牛之路 这一篇还是继续写react router相关的内容,废话之前先安利:https://github.com/ddwhan0123/Useful-Open-Source-Android (总有你须要的东西) 上一篇讲到我们能够利用 Router来实现嵌套跳转等效果,可是 ...
分类:其他好文   时间:2017-08-12 18:56:04    阅读次数:187
apache zookeeper的安装
original article:http://zookeeper.praveendeshmane.co.in/zookeeper/zookeeper-3-4-6-single-server-setup-on-ubuntu-14-04.jsp Apache ZooKeeper is a softwa ...
分类:Web程序   时间:2017-08-12 18:53:48    阅读次数:240
POJ - 3159 Candies
During the kindergarten days, flymouse was the monitor of his class. Occasionally the head-teacher brought the kids of flymouse’s class a large bag of ...
分类:其他好文   时间:2017-08-12 10:14:46    阅读次数:171
XYNU-ACM-ACboy needs your help again!
题目描述 ACboy was kidnapped!! he miss his mother very much and is very scare now.You can't image how dark the room he was put into is, so poor :(. As a s ...
分类:其他好文   时间:2017-08-11 22:58:19    阅读次数:142
好记性不如烂笔头77-多线程-Thread子类的线程对象是不同的
Thread子类的线程对象是不同的。 比方: EasySelfThread thread = new EasySelfThread(); //同一个线程对象 Thread t1 = new Thread(thread, “t1”); Thread t2 = new Thread(thread, “= ...
分类:编程语言   时间:2017-08-11 22:05:37    阅读次数:158
[bzoj1614]: [Usaco2007 Jan]Telephone Lines架设电话线
传送门 题意:给一个图,定义两点间的距离为路径上最大的边权,可以将路径上不多于k条边的权值变为0,求两点间最小距离 二分答案,判断时只要将大于当前二分值的边记为1,否则记为0,做一遍spfa,判断dist是否大于k即可 无向边数忘记乘2卡了半天T_T 1 #include<cstdio> 2 #in ...
分类:其他好文   时间:2017-08-11 20:28:49    阅读次数:184
SparkStreaming python 读取kafka数据将结果输出到单个指定本地文件
主要是重写pprint()函数 参考:https://stackoverflow.com/questions/37864526/append-spark-dstream-to-a-single-file-in-python ...
分类:编程语言   时间:2017-08-11 19:56:40    阅读次数:503
Divisibility by Eight
把当前数删除几位然后能够整除与8 那么可得知大于3位数的推断能否整除于八的条件是(n%1000)%8==0 能够得出我们的结论:仅仅须要枚举后三位后两位后一位就可以知道是否可整除于8 #include <cstdio> #include <cstring> #include <algorithm> ...
分类:其他好文   时间:2017-08-11 10:50:42    阅读次数:113
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