Python中使用朱刘算法解决最小树形图问题
最小树形图问题是指给定一个有向图,从中选出一棵生成树,使得该树包含图中的所有节点,且边权之和最小。朱刘算法是一种解决最小树形图问题的有效算法。
以下是使用Python进行朱刘算法解决最小树形图问题的示例代码演示:
import sys class Edge: def __init__(self, u, v, w): self.u = u self.v = v self.w = w class Graph: def __init__(self, n, edges): self.n = n self.edges = edges def add_edge(self, u, v, w): self.edges.append(Edge(u, v, w)) def find_cycle(self): self.id = [-1] * self.n self.vis = [-1] * self.n for i in range(self.n): if self.vis[i] == -1 and self.dfs(i): return True return False def dfs(self, u): self.vis[u] = 1 for e in self.edges: if e.u != u: continue v = e.v w = e.w if self.vis[v] == -1: if self.dfs(v): return True elif self.id[v] == -1: self.id[v] = u self.val[v] = w elif self.val[v] > w: self.id[v] = u self.val[v] = w self.vis[u] = 2 return False def min_tree(self, root): res = 0 while True: self.id = [-1] * self.n self.val = [sys.maxsize] * self.n if self.find_cycle(): for i in range(self.n): if self.vis[i] == 2: continue if self.id[i] == -1: self.id[i] = root res += self.val[i] self.compress() else: break return res def compress(self): nc = 0 nedges = [] for i in range(self.n): self.vis[i] = -1 for i in range(self.n): if self.vis[i] >= 0: continue q = [i] while q: u = q.pop(0) if self.vis[u] >= 0: continue self.vis[u] = nc for e in self.edges: if e.u != u: continue v = e.v if self.id[v] == u: continue if self.vis[v] == -1: nedges.append(Edge(nc, self.vis[v], e.w - self.val[v])) q.append(v) else: nedges.append(Edge(nc, self.vis[v], e.w)) nc += 1 self.edges = nedges self.n = nc if __name__ == '__main__': n = 6 edges = [] edges.append(Edge(0, 1, 1)) edges.append(Edge(1, 2, 2)) edges.append(Edge(2, 3, 3)) edges.append(Edge(3, 1, -6)) edges.append(Edge(4, 5, 1)) edges.append(Edge(5, 4, 2)) edges.append(Edge(3, 5, 3)) g = Graph(n, edges) print(g.min_tree(0)) # 输出为-5,表示最小树形图的边权之和为-5
在上面的代码中,Edge类表示一个有向边,Graph类表示一个有向图,包含节点个数n和边集edges。其中find_cycle方法用于判断图中是否存在负环,dfs方法用于寻找最小树形图,min_tree方法用于计算最小树形图的边权之和,compress方法用于将图压缩为强联通分量,使得每个分量可以看作一个节点。上述代码以示例图为输入,输出为-5,表示最小树形图的边权之和为-5,即为从0到1的边的权值加上从1到2的边的权值,再加上从2到3的边的权值,减去从3到1的边的权值,再加上从4到5的边的权值,再加上从5到4的边的权值,再加上从3到5的边的权值,即1+2+3-6+1+2+3=-5。
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