最长路径算法与递归方法的计算复杂度 [英] Computational complexity of a longest path algorithm witn a recursive method

问题描述

我写了一个代码段来确定图中的最长路径.以下是代码.但是由于中间有递归方法，我不知道如何获得计算复杂性.由于找到最长的路径是一个NP完整问题，因此我假设它类似于O(n!)或O(2^n)，但是我该如何确定呢?

I wrote a code segment to determine the longest path in a graph. Following is the code. But I don't know how to get the computational complexity in it because of the recursive method in the middle. Since finding the longest path is an NP complete problem I assume it's something like O(n!) or O(2^n), but how can I actually determine it?

public static int longestPath(int A) {
    int k;
    int dist2=0;
    int max=0;

    visited[A] = true;

    for (k = 1; k <= V; ++k) {
        if(!visited[k]){
            dist2= length[A][k]+longestPath(k);
            if(dist2>max){
                max=dist2;
            }
        }
    }
    visited[A]=false;
    return(max);
}

推荐答案

您的重复关系为T(n, m) = mT(n, m-1) + O(n)，其中n表示节点数，而m表示未访问节点数(因为您调用longestPath m次，并且有一个循环执行访问的测试n次).基本情况是T(n, 0) = O(n)(只是访问过的测试).

Your recurrence relation is T(n, m) = mT(n, m-1) + O(n), where n denotes number of nodes and m denotes number of unvisited nodes (because you call longestPath m times, and there is a loop which executes the visited test n times). The base case is T(n, 0) = O(n) (just the visited test).

解决这个问题，我相信您得到的T(n，n)是O(n * n！).

Solve this and I believe you get T(n, n) is O(n * n!).

编辑

工作:

T(n, n) = nT(n, n-1) + O(n) 
        = n((n-1)T(n, n-2) + O(n)) + O(n) = ...
        = n(n-1)...1T(n, 0) + O(n)(1 + n + n(n-1) + ... + n(n-1)...2)
        = O(n)(1 + n + n(n-1) + ... + n!)
        = O(n)O(n!) (see http://oeis.org/A000522)
        = O(n*n!)

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