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[Other] 动态规划----矩阵连乘问题 DL : 0: 动态规划----矩阵连乘问题动态规划法是解决问题的一种方法。它不规定为了得到结果需如何将问题划分为子问题的固定方法，而是按不同输入给出问题的具体实例的子问题划分方法，然后再进行运算、解答问题。矩阵连乘问题的主要思想如下： 1）设置大小为连乘个数的方阵 2）主对角线上方各元素Di,j(i<j)表示矩阵Mi连乘到Mj的最小工作量 3）下方元素Di,j(i>j)记录获得该最小工作量矩阵分组的第一组的最后一个矩阵的序列号最后通过下方元素可知最终结果的分组方式。-dynamic programming matrix continually multiply-dynamic programming problem is a problem-solving method. It does not require the results need to be how to divide the problems of the sub- problems fixed, but different input given by the specific example of the problem partition method, and then calculate, and answer questions. Matrix continually multiply the main idea is as follows : 1) the installation of the size of continually multiply the number phalanx 2) above the main diagonal elements Di, j (ilt; J) Matrix Mi continually multiply to the smallest workload 3) below elements Di, j (IGT; J) the record was the smallest workload of a matrix of the first group of a matrix of the final sequence, followed by the final element of the final results of the known clusters.
Date : 2026-01-02 Size : 26kb User : 莫非
[Other] 23456787654321 DL : 0: 问题一：设a[0:n-1]是已排好序的数组。请改写二分搜索算法，使得当搜索元素x不在数组中时，返回小于x的最大元素位置i和大于x的最小元素位置j，当搜索元素在数组中时，i和j相同，均为x在数组中的位置。问题二：采用分治策略实现对n个元素进行排序的合并排序算法，其基本思想是：将待排序元素分成大小大致相同的2个子集合，分别对2个子集合进行排序，最终将排好序的子集合合并成为所要求的排好序的集合。问题三：给定n个矩阵{A1，A2，A3，……，An}，其中，Ai与Ai+1是可乘的，i=1，2，…,n-1。要求用动态规划解矩阵连乘积的最优计算次序问题。该三题的算法设计思想及源码，调试结果-Question one: Let a [0: n-1] is already sorted array. Please rewrite the binary search algorithm so that when the search element of x is not an array, the return is less than x, is larger than the largest element in position i and position x, the smallest element of j, when the search elements in the array when, i and j the same, both x in the array position. Problem 2: The strategy for achieving the partition of n elements to sort of merge sort algorithm, whose basic idea is: sort elements are to be divided into roughly the same size of the two sub-sets, respectively, of the two sub-sort the collection will eventually sorted The merged to form the required subset sorted collection. Question 3: Given n matrices (A1, A2, A3, ... ..., An), where, Ai and Ai+1 is a mere a, i = 1,2, ..., n-1. Required to use dynamic programming solution matrix with the optimal product sequence calculation of the problem. The three questions of the algorithm design ideas and source code, debu
Date : 2026-01-02 Size : 9kb User : 土咩豆
[Other] MCSP DL : 0: 计算两个字符串的最小公共子串划分（MCSP）。最小公共子串划分（MCSP)是指把两个字符串划分成相同的子串的集合，同时要求划分的子串数目最少。比如，S1 = cdabcdabceab，S2 = abceabcdabcd,求得的MCSP是(cdabcd, abceab), (abceab, cdabcd).MCSP问题是基因重组中的关键问题，并且与用逆转重复字串进行排序的问题十分相近。MCSP问题是NP难的。-The code is to compute MCSP of two given strings. The Minimum Common String Partition problem (MCSP) is to partition two given input strings into the same collection of substrings, where the number of substrings in the partition is minimized. This problem is a key problem in genome rearrangement, and is closely related to the problem of sorting by reversals with duplicates. MCSP is NP-hard, even for the most trivial case, 2-MCSP, where each letter occurs at most twice in each input string.
Date : 2026-01-02 Size : 3kb User : luckypig
[Other] fenqu DL : 0: 电脑分区时是否经常面临分区容量不整齐的问题？这个小程序能让你的分区很整齐。-Whether the computer partition partition size irregular often face the problem? This little program allows you to partition a very tidy.
Date : 2026-01-02 Size : 6kb User : linhong
[Other] ACM DL : 0: 本文包含本人亲自写的两个ACM的小算法，一个是石子划分问题另一个是服务器转移问题，全是用的是动态规划算法，经试验，这两个算法运行效率极高-This article contains personally wrote two small ACM algorithm, a stone partition problem is transferred to another is a server problem, all using a dynamic programming algorithm, and tested, these two very efficient algorithm runs
Date : 2026-01-02 Size : 1kb User : 110
[Other] abc DL : 0: 投掷六个骰子能投掷出多少种排列组合呢？如1 2 3 4 5 6 和 6 5 4 3 2 1是同一种组合。问题可以转化为：n个相同小球放入r个相异盒子中，允许空盒。即求解不定方程 x1+x2+x3+ ... + xr = n 的非负整数解(x1, x2, x3, ..., xr)，0<= Xi <= n 令yi=xi+1,转化为 y1+y2+..+yr = n+r 此时 1<= yi <=n+1 相当于 n+r 个小球分成 r 堆，有多少种分法的问题即在这n+r-1个间隔中放置r-1个隔板，隔板之间的球的个数就相当于yi. 答案是 C(n+r-1,r-1) 本程序利用VC++语言解决本问题。-Throw six dice can throw out many permutations of it? As 123456 and 654321 is the same composition. Problem can be transformed into: n r identical balls into a different box, allowing the empty boxes. That solved indefinite equation x1+x2+x3+ ...+ xr = n of non-negative integer solutions (x1, x2, x3, ..., xr), 0 < = Xi < = n so yi = xi+1, into y1+y2+ ..+yr = n+r this case 1 < = yi < = n+1 equivalent n+r r balls into the heap, there are many kinds of issues of law that r-1 partition plates placed in this n+r-1 intervals, the ball separator between The number is equivalent to yi. answer is C (n+r-1, r-1) This procedure using VC++ solve the language problem.
Date : 2026-01-02 Size : 190kb User : 王涛
[Other] voronoi DL : 0: Voronoi Diagram算法实现主要运用分治法和凸包的方法来实现。分治法： 1) 该问题的规模缩小到一定的程度就可以容易地解决 2) 该问题可以分解为若干个规模较小的相同问题，即该问题具有最优子结构性质。 3) 利用该问题分解出的子问题的解可以合并为该问题的解； 4) 该问题所分解出的各个子问题是相互独立的，即子问题之间不包含公共的子问题。 -Voronoi Diagram algorithm is mainly using the method of partition and the method of convex hull. Divide and conquer method: 1) reduced to a certain degree the problem can be easily solved 2) the problem can be broken down into several smaller same problem, that is, the problem is in the nature of optimal substructure. 3) by using the solution of problem into sub-problems can be combined for the solution of the problem 4) the decomposition of the problem each subproblem is independent of each other, between the subproblems do not contain public subproblems.
Date : 2026-01-02 Size : 3kb User : 李阳
[Other] solution-for-hannuota DL : 0: 汉诺塔解法代码,可以解决任意层数的汉诺塔问题，主要方法是递归与分治的思想-Tower of Hanoi solution code, you can solve any number of layers of the Tower of Hanoi problem, the main method is recursive with the idea of partition
Date : 2026-01-02 Size : 3.76mb User : bujingyun
[Other] integer-partitioning DL : 0: 整数划分问题编程序求某一个正整数的所有划分数-Integer partition problem all programmed to seek a number of divisions a positive integer.
Date : 2026-01-02 Size : 1kb User : Adrian