最小费用最大流问题的二分图最小权匹配解法!(数据满足一定条件:二分图,拆点数较小)。...
分类:
其他好文 时间:
2014-12-06 19:33:17
阅读次数:
220
Given two wordsword1andword2, find the minimum number of steps required to convertword1toword2. (each operation is counted as 1 step.)You have the fol...
分类:
其他好文 时间:
2014-12-06 16:46:02
阅读次数:
202
1.width=device-width //应用程序的宽度和屏幕的宽度是一样的2.height=device-height //应用程序的高度和屏幕的高是一样的3.initial-scale=1.0 //应用程序启动时候的缩放尺度(1.0表示不缩放)4.minimum-scale=1.0 //用户...
分类:
移动开发 时间:
2014-12-06 13:59:57
阅读次数:
244
Triangle Given a triangle, find the minimum path sum from top to bottom. Each step you may move to adjacent numbers on the row below.For example, give...
分类:
其他好文 时间:
2014-12-06 06:33:39
阅读次数:
186
Given a triangle, find the minimum path sum from top to bottom. Each step you may move to adjacent numbers on the row below.
For example, given the following triangle
[
[2],
[3,4],
[...
分类:
其他好文 时间:
2014-12-05 22:50:19
阅读次数:
151
1 /*Find Minimum in Rotated Sorted Array */ 2 #include 3 #include 4 using namespace std; 5 6 int find(vector & num, int begin, int end) ; 7 int fi...
分类:
其他好文 时间:
2014-12-05 19:07:46
阅读次数:
204
这里简单加上几个验证,非空,最小长度,唯一修改模型修改app/models/post.rb文件,如下:class Post true validates :title, :length => { :minimum =>2} validates :title, :uniqueness => { ...
分类:
其他好文 时间:
2014-12-04 17:08:31
阅读次数:
112
Design a stack that supports push, pop, top, and retrieving the minimum element in constant time.
push(x) -- Push element x onto stack.pop() -- Removes the element on top of the stack.top() -- Get ...
分类:
其他好文 时间:
2014-12-04 12:21:07
阅读次数:
128
In this problem you are asked to convert a string into a palindrome with minimum number of operations. The operations are described below:
Here you’d have the ultimate freedom. You are allowed to:...
分类:
其他好文 时间:
2014-12-04 01:03:49
阅读次数:
138
Given two wordsword1andword2, find the minimum number of steps required to convertword1toword2. (each operation is counted as 1 step.)You have the fol...
分类:
其他好文 时间:
2014-12-03 22:42:21
阅读次数:
130