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poj 3625 Building Roads
Building Roads Time Limit: 1000MS   Memory Limit: 65536K Total Submissions: 8999   Accepted: 2596 Description Farmer John had just acquired several new farms! He wants to...
分类:其他好文   时间:2014-07-21 23:29:43    阅读次数:232
HDU 4584 Building bridges
枚举。 又是一期检测题。又是只能做水题。唉。为毛老是选这种邀请赛的题。 枚举一下 H 点 和 C 点就行了。 #include #include #include #include #include #include #include #include #include #include #include #define INF 0x7fffffff #define eps ...
分类:其他好文   时间:2014-07-21 15:33:15    阅读次数:200
UVA101 The Blocks Problem
思路就是用栈来模拟,一开始就是n个栈。每个栈里都是一个元素,然后按照指令移,在这个栈里pop()掉它,在另一个栈里push()进去。。 分四种情况来做移动,每种情况处理方式不一样。要注意如果是一堆移过去,因为还是要按照这个顺序,多以要先把这一堆放到另一个数组,再按顺序pushj进去。 模拟完输出即可。。...
分类:其他好文   时间:2014-07-21 15:11:15    阅读次数:212
Swift_4_闭包(Blocks)
import Foundation println("Hello, World!") var arr = [1,2,4,6,74,2] func hasClosure(list:[Int], v2:Int, cb:(num:Int, v3:Int) ->Bool) ->Bool{ for item in arr{ if(cb(num:item, v3:v2...
分类:其他好文   时间:2014-07-21 13:15:37    阅读次数:250
UVa 825 Walking on the Safe Side(DP)
Square City is a very easy place for people to walk around. The two-way streets run North-South or East-West dividing the city into regular blocks. Most street int...
分类:其他好文   时间:2014-07-20 22:33:53    阅读次数:295
Easy and cheap cluster building on AWS backup
https://grapeot.me/easy-and-cheap-cluster-building-on-aws.htmlThu 17 July 2014, byYan Wang|2 CommentsLinuxParallelgithubImageWhy?It often requires a l...
分类:其他好文   时间:2014-07-19 23:09:27    阅读次数:310
构建针对 iOS 和 Android 的原生扩展
第一部分:音量控制入门http://www.adobe.com/cn/devnet/air/articles/building-ane-ios-android-pt1.html第二部分: 开发 ActionScript 库http://www.adobe.com/cn/devnet/air/arti...
分类:移动开发   时间:2014-07-19 16:00:23    阅读次数:289
[POJ 1390]Blocks
Description Some of you may have played a game called 'Blocks'. There are n blocks in a row, each box has a color. Here is an example: Gold, Silver, Silver, Silver, Silver, Bronze, Bronze, Bronze, Go...
分类:其他好文   时间:2014-07-18 11:23:05    阅读次数:280
USACO cowtour Floyd + 枚举
给出来的数据量还是可以的。题意:有若干个牧场,至少有两个不连通,一个牧场的直径就是牧场中最远的两个牧区的距离。要求找出几个牧场中最短的直径,就是找一条路径连接几个牧区,使这个直径最终最小。 基本方法,把整个图根据输入划分成几个不连通的牧区,然后求出每个牧区的直径(即每个连通块中的最长路径),然后枚举两个不在同一牧区的点,设blocks[i]记录第i个节点所在连通块的直径,那么result = m...
分类:其他好文   时间:2014-07-18 11:04:46    阅读次数:287
POJ 3566
?? Building for UN Time Limit: 5000MS   Memory Limit: 65536K Total Submissions: 1367   Accepted: 599   Special Judge Description The United Nations has decided to buil...
分类:其他好文   时间:2014-07-17 20:37:19    阅读次数:294
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