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  • 最小生成树(Minimum Cost Spanning Tree)2020-01-28 10:37:35

    MST的性质 若(u, v)是一条具有最小权值的边,则存在一棵包含边(u, v)的最小生成树 最小生成树的算法就是基于MST这个性质写的 一、Prim算法 Prim算法也称为"加点法",从v0开始,根据MST的性质每次添加一条最小花费的顶点到集合U Prim特有属性 class Closedge { // closedge[]的

  • 高光谱图像处理---linearly constrained minimum variance线下约束最小方差(LCMV)2020-01-27 14:01:16

    linearly constrained minimum variance线性约束最小方差(LCMV) 写在前面:为了更好的后期修改,如果各位觉得有哪里可以补充或者优化,直接从我这里下载,然后发给我,你也是作者之一。 本文文本下载链接://download.csdn.net/download/qq_43110298/12116806(也可以在本人的下载中找到) 直

  • 【WPF学习】第十二章 属性验证2020-01-24 19:54:32

      在定义任何类型的属性时,都需要面对错误设置属性的可能性。对于传统的.NET属性,可尝试在属性设置器中捕获这类问题。但对于依赖项属性而言,这种方法不合适,因为可能通过WPF属性系统使用SetValue()方法直接设置属性。   作为代替,WPF提供了两种方法来阻止非法值: ValidateValueCal

  • [LeetCode 1167] Minimum Cost to Connect Sticks2020-01-24 09:02:29

    You have some sticks with positive integer lengths. You can connect any two sticks of lengths X and Y into one stick by paying a cost of X + Y.  You perform this action until there is one stick remaining. Return the minimum cost of connecting all t

  • Triangle2020-01-21 16:08:38

    120. 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, given the following triangle [ [2], [3,4], [6,5,7], [4,1,8,3] ] The minimum path sum f

  • LeetCode 727. Minimum Window Subsequence2019-12-28 14:02:27

    原题链接在这里:https://leetcode.com/problems/minimum-window-subsequence/ 题目: Given strings S and T, find the minimum (contiguous) substring W of S, so that T is a subsequence of W. If there is no such window in S that covers all characters in T

  • Minimum Path Sum2019-12-21 21:03:59

    Description Given a m x n grid filled with non-negative numbers, find a path from top left to bottom right which minimizes the sum of all numbers along its path. You can only go right or down in the path.. Example Example 1: Input: [[1,3,1],[1,

  • Minimum Size Subarray Sum2019-12-21 13:53:45

    Given an array of n positive integers and a positive integer s, find the minimal length of a contiguous subarray of which the sum ≥ s. If there isn't one, return 0 instead.     用个比较笨的方法来做,因为我本人目前并没有能力想到啥高深的算法来解决此问题。(废话)

  • 【AGC030F】Permutation and Minimum(DP)2019-12-08 14:04:22

    题目链接 题解 首先可以想到分组后,去掉两边都填了数的组。 然后就会剩下\((-1,-1)\)和\((-1,x)\)或\((x,-1)\)这两种情况 因为是最小值序列的情况数,我们可以考虑从大到小填数,一个组填完了最小值就是当前填的数 设\(f[i][j][k]\)表示已经填到数\(i\)剩余\(j+k\)个填了一个数的组,其

  • [思维]Minimum Spanning Tree2019-10-27 19:54:24

    题目描述 In the mathematical discipline of graph theory, the line graph of a simple undirected weighted graph G is another simple undirected weighted graph L(G) that represents the adjacency between every two edges in G.Precisely speaking, for an undirecte

  • Design a stack that supports getMin() in O(1) time and O(1) extra space2019-10-22 10:05:24

    Question: Design a Data Structure SpecialStack that supports all the stack operations like push(), pop(), isEmpty(), isFull() and an additional operation getMin() which should return minimum element from the SpecialStack. All these operations of SpecialSt

  • Python3程序设计指南:01 过程型程序设计快速入门2019-10-16 20:56:14

    大家好,从本文开始将逐渐更新Python教程指南系列,为什么叫指南呢?因为本系列是参考《Python3程序设计指南》,也是作者的学习笔记,希望与读者共同学习。 .py文件中的每个估计都是顺序执行的,从第一行开始,逐行执行的。 Python的8个关键要素 1、数据类型 int整数(正整数或负整数) Py

  • 知识总结2019-10-12 20:54:21

    #知识总结 ##一.思维导图 ##二.if-else;switch;for;while;do-while; ###break;continue; *1. if(条件){语句 } else{ 语句 } ​ 常见问题:1. “{}”的书写不完整。2. 对于else附属于哪个if不确定。3. 解决方法:在本地运行代码。 switch(表达式) { case 常量表达式1:语句段1;break; ​

  • php-如何从数组行中获取最小值2019-10-12 14:32:39

    我试图从列名称中包含“ xx”的任何列中获取最小值. 下面是我的代码: <?php $array = array( array( 'id' => 1, '10xx' => 14, '11xx' => 32, '12xx' => 4 ), array( 'id' => 2, '10xx' => 13, &#

  • javascript – 查找两个数组之间的常见最小值2019-10-07 13:38:37

    我正在研究Javascript中的问题.查找两个数组之间的公共最小值.但是,有人告诉我,这可能不适用于某些值.有什么问题? function cmp(a, b) { return a - b; } function findMinimum(A, B) { var n = A.length; var m = B.length; A.sort(cmp); B.sort(cmp); var

  • LeetCode.1200-最小绝对值差(Minimum Absolute Difference)2019-10-06 21:03:02

    这是小川的第418次更新,第451篇原创 看题和准备 今天介绍的是LeetCode算法题中Easy级别的第268题(顺位题号是1200)。给定一个由不同的整数组成的数组arr,找到所有对元素,其中任意两个元素的绝对差值都最小。 以升序返回关于配对的列表(相对于配对),每对[a,b]紧随其后: a,b来自arr a < b b-a

  • AT2060 【Minimum Sum】2019-09-25 17:54:19

    单调栈 这是某次olinr巨佬给我们出的考试题。首先暴力\(O(n^2)\)是不能过的(废话),我们考虑每一个数值对答案的贡献,也就是ta能当最小值的序列个数。 倘若ta能成为最小值,那么也就是在这个数列里没有比ta大的数值,这也就转化为了求出右边第一个比ta小的值的位置和左边第一个比ta小的值

  • Leetcode 64 Minimum Path Sum2019-09-21 22:01:15

    Minimum Path Sum Given a m x n grid filled with non-negative numbers, find a path from top left to bottom right which minimizes the sum of all numbers along its path. Note: You can only move either down or right at any point in time. Example: Input: [

  • Aggressive cows (北京大学ACM-ICPC竞赛训练暑期课 )2019-09-19 23:02:57

    描述Farmer John has built a new long barn, with N (2 <= N <= 100,000) stalls. The stalls are located along a straight line at positions x1,...,xN (0 <= xi <= 1,000,000,000). His C (2 <= C <= N) cows don't like this barn layout and be

  • Minimum Height Trees 解答2019-09-13 22:57:09

    Question For an undirected graph with tree characteristics, we can choose any node as the root. The result graph is then a rooted tree. Among all possible rooted trees, those with minimum height are called minimum height trees (MHTs). Given such a graph,

  • 【Java】【LeetCode】64. Minimum Path Sum2019-08-28 17:01:34

    题目: Given a m x n grid filled with non-negative numbers, find a path from top left to bottom right which minimizes the sum of all numbers along its path. Note: You can only move either down or right at any point in time. Example: Input: [   [1,3

  • 在Java中使用Double.POSITIVE_INFINITY来查找最小值2019-08-25 14:10:26

    简单的问题:以下代码是否可用于查找双精度数组中的最小值(假设至少存在一个值): double[] values = ... double currentMin = Double.POSITIVE_INFINITY; for(int i = 0; i < values.length; i++) { if(values[i] < currentMin) { currentMin = values[i]; } } r

  • 实例49:python2019-08-19 12:36:02

    #题目:使用lambda来创建匿名函数。 #!/usr/bin/python -- coding: UTF-8 -- MAXIMUM = lambda x,y : (x > y) * x + (x < y) * y MINIMUM = lambda x,y : (x > y) * y + (x < y) * x if name == ‘main’: a = 10 b = 20 print (‘The largar one is %d’ % MAXIMUM(a,b))

  • 实例432019-08-14 09:44:10

    #include<stdio.h> #define MAX #define MAXIMUM(x,y)(x>y)?x:y #define MINIMUM(x,y)(x>y)?y:x int main() { int a=10,b=20; #ifdef MAX printf(“更大的数字是 %d\n”,MAXIMUM(a,b)); #else printf(“更小的数字是 %d\n”,MINIMUM(a,b)); #endif #ifndef MIN printf(“

  • 2019牛客暑期多校训练营(第二场)----D-Kth Minimum Clique2019-08-04 16:02:00

    首先发出题目链接: 链接:https://ac.nowcoder.com/acm/contest/882/D 来源:牛客网 涉及:优先队列 点击这里回到2019牛客暑期多校训练营解题—目录贴 题目如下:

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