Sorting

题目链接: http://acm.hdu.edu.cn/showproblem.php?pid=5122 Problem Description Matt's friend K.Bro is an ACMer. Yesterday, K.Bro learnt an algorithm: Bubble sort. Bubble sort will compare each pair of adjacent items and swap them if they are in the wrong o

You are given a sequence a1,?a2,?...,?an consisting of different integers. It is required to split this sequence into the maximum number of subsequences such that after sorting integers in each of them in increasing order, the total sequence also wil

1,layer是对游戏中所有物体的分类别划分,如UIlayer, waterlayer, 3DModelLayer, smallAssetsLayer, effectLayer等.将不同类的物体划分到不同的层,便于相机拣选,在相机的cullmask中可以选择渲染哪些层,不选择的层则不会渲染.还可以用于射线检测对象的拣选,可以指定只对某些层的对象进行射线检测. 2,canvas默认是屏幕空间的2D对象,在屏幕空间时仅具有sort order属性,当把它设置为世界空间时,sort order属性消失

DZY Loves Topological Sorting Time Limit: 4000/2000 MS (Java/Others) Memory Limit: 131072/131072 K (Java/Others) Total Submission(s): 922 Accepted Submission(s): 269 Problem Description A topological sort or topological ordering of a directed graph i

hdu2838 ------希望30号驾校科目一顺利考完,4月即将过去说好的30篇博客也没完成, 真是忙起来就会烦躁什么都不想做,勿忘心安.... <Cow Sorting> 这题本来兴高采烈的想用java做一遍,结果做出来之后无限超内存,真是啊,做题的时候java这种东西还是轻易不要动了.还有感觉要把数字都要离散化的,结果后台数据不需要离散化. 题意:给一个n代表n个牛,然后再给n个数我觉得是n以内的数(包括n).虽然体面上没说.然后只能相邻两个数字交换位置,会让牛产生怒气值,值为互相移动的

K.Bro Sorting Time Limit: 2000/2000 MS (Java/Others)    Memory Limit: 512000/512000 K (Java/Others) Total Submission(s): 67    Accepted Submission(s): 39 Problem Description Matt's friend K.Bro is an ACMer. Yesterday, K.Bro learnt an algorithm: Bubbl

WebGrid with filtering, paging and sorting by Jose M. Aguilar on April 24, 2012 in Web Development A few days ago I received some questions on the use of the Webgrid helper in the comments section of my personal blog, specifically on how to implement

ASP.NET MVC WebGrid – Performing true AJAX pagination and sorting FEBRUARY 27, 2012 14 COMMENTS WebGrid is a very powerful HTML helper component introduced with ASP.NET MVC 3 and ASP.NET Web Pages. Despite all its cool features, it is troublesome to

Time complexity: Binary search O(log2 n): i=0.   n elements:         ------------------- i=1.   n/2 elements:                   ---------- i=2.   n/4 elements:                          ----- ... i=i.    n/2^i elements:                        - 进行n/2^

题目链接 题意:给出n个数,求出得到最大数和第二大数所用的最多的比较次数 思路:可以取两个数两两做比较,就相当与建立二叉树,两个数的最大值就相当与两个的父节点,所以我们我们可以知道得到最大值时的比较次数为n-1,然后假设所有左儿子都大于右儿子,所以最大值的右儿子还有跟左子树中的值做比较得到第二大数,比较的个数为树高. 代码: #include <iostream> #include <cstdio> #include <cstring> #include <cma