In this article, we will use C to count the number of paths with weight W in the K-ary tree. We have been given a K-ary tree, which is a tree in which each node has K children and each edge has a weight, with the weight decreasing from 1 to K from a node to all its children.
We need to count the cumulative number of paths starting from the root node that have weight W and at least one edge with weight M. So, here is an example:
Input : W = 4, K = 3, M = 2 Output : 6
In the given problem, we will use dp to reduce the time and space complexity. By using memoization, we can make our programs faster and adapt them to larger constraints.
Method
In this method we will traverse the tree and keep track of edges with or without weight of at least M and weight equal to W, then we increment the answer.
Input
#include <bits/stdc++.h> using namespace std; int solve(int DP[][2], int W, int K, int M, int used){ if (W < 0) // if W becomes less than 0 then return 0 return 0; if (W == 0) { if (used) // if used is not zero then return 1 return 1; //as at least one edge of weight M is included return 0; } if (DP[W][used] != -1) // if DP[W][used] is not -1 so that means it has been visited. return DP[W][used]; int answer = 0; for (int i = 1; i <= K; i++) { if (i >= M) answer += solve(DP, W - i, K, M, used | 1); // if the condition is true //then we will change used to 1. else answer += solve(DP, W - i, K, M, used); } return answer; } int main(){ int W = 3; // weight. int K = 3; // the number of children a node has. int M = 2; // we need to include an edge with weight at least 2. int DP[W + 1][2]; // the DP array which will memset(DP, -1, sizeof(DP)); // initializing the array with -1 value cout << solve(DP, W, K, M, 0) << "\n"; return 0; }
Output
3
Explanation of the above code
In this method, edges with weight M are included at least once or not . Second, we calculated the total weight of the path if it is equal to W.
We increment the answer by one, mark the path as visited, continue through all possible paths, and contain at least one edge with weight greater than or equal to M.
Conclusion
In this article, we use dynamic programming to solve the problem of finding the number of paths with weight W in a k-ary tree, with a time complexity of O(W*K ).
We also learned the C program and the complete method (common and efficient) to solve this problem.
The above is the detailed content of Find the number of paths with weight W in a K-ary tree using C++. For more information, please follow other related articles on the PHP Chinese website!

微博权重是指微博官方对微博号的评分,主要体现在搜索和评论时的排序,权重越高排序越靠前,因此微博权重也会影响到微博号的流量数据。提高权重可以通过实名制的方式,还可以成为微博的签约自媒体。

深入理解CSS选择器通配符的权重和优先级在CSS样式表中,选择器是用来指定样式应用于哪些HTML元素的重要工具。选择器的优先级和权重决定了当多个规则同时作用于一个HTML元素时,应用哪个样式。通配符选择器是CSS中一种常见的选择器。它使用“*”符号表示,表示匹配所有HTML元素。通配符选择器虽然简单,但在某些情况下非常有用。然而,通配符选择器的权重和优先级也

在本文中,给定的任务是找到字符串的总重量。为了计算字符串权重,我们将给定的字符串转换为较低的形式。考虑到字符的重量,我们取a=1、b=,2等等,直到z=26。在这篇Python文章中,使用两个不同的示例,给出了查找给定字符串的权重的方法。在第一个示例中,字符串中的给定字符为fetc、hed,然后将它们各自的权重添加到更新后的权重中。在示例2中,首先计算给定字符在字符串中出现的频率,然后将该频率乘以相应的字符权重,然后将所有这些分量权重加在一起得到最终结果。示例1:使用迭代查找字符串权重并添加字符

在抖音这一备受瞩目的短视频平台上,拥有一个高权重的账号是众多用户梦寐以求的。然而,对于一些低权重的账户来说,如何提升抖音权重成为了一个亟需解决的问题。本文旨在探讨低权重账号的提升之道,分享一些有效的方法和技巧。1、提供优质内容:无论在什么平台上,内容始终是最重要的。为了提升抖音账号的权重,你需要提供具有吸引力和独特性的优质内容。这意味着你应该创作出有趣、有价值和与众不同的视频,能够引起观众的兴趣和共鸣。关注热门话题和流行潮流,不断创新和尝试新的创意,吸引更多的观众关注和点赞。2、与观众互动:积极

为了发现具有权重大于或等于1的最小边的路径,我们可以使用稍作修改的Dijkstra算法。首先,我们将源节点的权重设为1,将其他节点的权重设为无穷大。在算法执行过程中,我们不再更新距离,而是更新权重的乘积。这样可以确保选择具有最小权重的路径。通过在每一步选择权重最小的节点,我们迭代地发现最短路径,直到达到目标节点。最后,沿着这条路径的权重乘积将是最小的,满足给定的条件。使用的方法修改后的Dijkstra算法,带有加权乘积修改的Bellman-Ford算法,带有加权乘积加权乘积的改进Dijkstra

在本文中,我们将使用C++来计算K叉树中权重为W的路径数量。我们已经给出了一个K叉树,它是一棵每个节点有K个子节点且每条边都有一个权重的树,权重从1到K递减从一个节点到其所有子节点。我们需要计算从根节点开始的累积路径数量,这些路径具有权重为W且至少有一条边的权重为M。所以,这是一个例子:Input:W=4,K=3,M=2Output:6在给定的问题中,我们将使用dp来减少时间和空间复杂度。通过使用记忆化,我们可以使我们的程序更快,并使其适用于更大的约束。方法在这个方法中,我们将遍历树,并跟踪使用

抖音作为目前最热门的社交媒体平台之一,拥有数以亿计的用户。然而,对于许多抖音创作者来说,他们面临着一个共同的问题,那就是抖音权重低。抖音权重低意味着他们的视频很难被推荐给更多的用户,从而影响了他们的曝光和粉丝增长。那么,面对这个问题,我们应该如何提升自己的抖音权重呢?一、抖音权重低怎么提升?关键词优化是提升抖音视频权重的关键。在发布视频的时候,我们要注意选择适合的关键词,这样有利于视频被搜索和推荐。可以通过调研流行的关键词和话题,找到与自己内容相关的关键词,并在标题、描述、标签中合理运用。还可以

在本文中,我们给出了一个问题,我们需要找到从点A到点B的总路径数,其中A和B是固定点,即A是网格中的左上角点,B是网格中的右下角点,例如−Input:N=5Output:252Input:N=4Output:70Input:N=3Output:20在给定的问题中,我们可以通过简单的观察来形式化答案并得出结果。寻找解决方案的方法在这种方法中,我们通过观察得出一个公式,即从A到B穿过网格时,我们需要向右行进n次,向下行进n次,这意味着我们需要找到所有可能的路径组合,因此我们得到了


Hot AI Tools

Undresser.AI Undress
AI-powered app for creating realistic nude photos

AI Clothes Remover
Online AI tool for removing clothes from photos.

Undress AI Tool
Undress images for free

Clothoff.io
AI clothes remover

AI Hentai Generator
Generate AI Hentai for free.

Hot Article

Hot Tools

Dreamweaver CS6
Visual web development tools

SecLists
SecLists is the ultimate security tester's companion. It is a collection of various types of lists that are frequently used during security assessments, all in one place. SecLists helps make security testing more efficient and productive by conveniently providing all the lists a security tester might need. List types include usernames, passwords, URLs, fuzzing payloads, sensitive data patterns, web shells, and more. The tester can simply pull this repository onto a new test machine and he will have access to every type of list he needs.

MantisBT
Mantis is an easy-to-deploy web-based defect tracking tool designed to aid in product defect tracking. It requires PHP, MySQL and a web server. Check out our demo and hosting services.

mPDF
mPDF is a PHP library that can generate PDF files from UTF-8 encoded HTML. The original author, Ian Back, wrote mPDF to output PDF files "on the fly" from his website and handle different languages. It is slower than original scripts like HTML2FPDF and produces larger files when using Unicode fonts, but supports CSS styles etc. and has a lot of enhancements. Supports almost all languages, including RTL (Arabic and Hebrew) and CJK (Chinese, Japanese and Korean). Supports nested block-level elements (such as P, DIV),

ZendStudio 13.5.1 Mac
Powerful PHP integrated development environment
