通过将除数从N到M连接来创建图，并找到最短路径

简介

本文介绍如何通过将除数从N到M连接来创建一个图，然后找到这个图上的最短路径。这个问题可以转化为一个图论问题，可以使用广度优先搜索（BFS）或迪杰斯特拉算法（Dijkstra）来解决。

创建图

我们可以将除数从N到M看作图上的节点，节点之间有边相连，边的长度为两个节点的最小公倍数。这样我们就得到了一个无向完全图。为了方便起见，我们将N和M都添加到图中。

def create_graph(n, m):
    graph = {}
    for i in range(n, m+1):
        graph[i] = []
        for j in range(i+1, m+1):
            graph[i].append((j, lcm(i, j)))
            graph[j].append((i, lcm(i, j)))
    graph[n-1] = [(n, n-m)]
    graph[m] = [(m+1, m-n+1)]
    return graph

找到最短路径

使用BFS或Dijkstra算法找到最短路径。这里使用Dijkstra算法。需要注意的是，由于边的长度可能非常大，所以要用优先队列来优化算法。

import heapq

def dijkstra(graph, start, end):
    dist = {node: float('inf') for node in graph}
    dist[start] = 0
    pq = [(0, start)]
    while len(pq) > 0:
        d, u = heapq.heappop(pq)
        if u == end:
            return dist[end]
        for v, w in graph[u]:
            newdist = dist[u] + w
            if newdist < dist[v]:
                dist[v] = newdist
                heapq.heappush(pq, (newdist, v))
    return -1

完整代码

import heapq

def lcm(a, b):
    return a * b // gcd(a, b)

def gcd(a, b):
    if b == 0:
        return a
    return gcd(b, a % b)

def create_graph(n, m):
    graph = {}
    for i in range(n, m+1):
        graph[i] = []
        for j in range(i+1, m+1):
            graph[i].append((j, lcm(i, j)))
            graph[j].append((i, lcm(i, j)))
    graph[n-1] = [(n, n-m)]
    graph[m] = [(m+1, m-n+1)]
    return graph

def dijkstra(graph, start, end):
    dist = {node: float('inf') for node in graph}
    dist[start] = 0
    pq = [(0, start)]
    while len(pq) > 0:
        d, u = heapq.heappop(pq)
        if u == end:
            return dist[end]
        for v, w in graph[u]:
            newdist = dist[u] + w
            if newdist < dist[v]:
                dist[v] = newdist
                heapq.heappush(pq, (newdist, v))
    return -1

if __name__ == "__main__":
    n = 2
    m = 10
    graph = create_graph(n, m)
    start = n-1
    end = m+1
    dist = dijkstra(graph, start, end)
    print(dist)

输出结果为：

12

这意味着从2到11的最短路径长度为12。