ADA算法知识（十）动态规划相应的问题

最新推荐文章于 2024-04-09 19:34:21 发布

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最新推荐文章于 2024-04-09 19:34:21 发布

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本文链接：https://blog.youkuaiyun.com/qq_42615643/article/details/85246843

本文探讨了如何使用动态规划解决在一个无向图中，从节点s到节点t，利用k条边的所有可能路径数量的问题。文章首先定义了问题的动态规划状态P(u,v,i)，然后给出了P(u,v,1)的初始条件，接着提出了递归公式，并讨论了使用Floyd-Warshall算法进行计算的方法，最后分析了算法的时间复杂度为O(n^3)。

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[How Many Paths?] You are given a map in the form of an undirected graph G = (V,E) and would like to know how many possible ways to walk (that are, paths with possible repeated nodes) between two nodes s and t, using k number of edges. You would like to use dynamic programming to solve the problem.

For any nodes u,v belongs to V, and i>=0, let P(u,v,i) denote the number of walks that starts from u and ends at v that use i edges. Note that question now is to compute P(s,t,k).

(a) What is the valu of P(u,v,1) for any u,v belong to V?

(b) Write down a recurrence for the value of P(u,v,i+1).

(a)Obviously, when i=1, P(u,v,1)=1

(b)use dfs

(c)use floydwarshall algorithm

and the complexity of the algorithm is O(n^3)