问题

描述：有A、B、C、D四个点，每两个点之间的距离（无方向）是(第一个数字是两点之间距离，后面两个字母代表两个点）：(1,‘A‘,‘B‘),(5,‘A‘,‘C‘),(3,‘A‘,‘D‘),(4,‘B‘,‘C‘),(2,‘B‘,‘D‘),(1,‘C‘,‘D‘) 生成边长和最小的树，也就是找出一种连接方法，将各点连接起来，并且各点之间的距离和最小。

思路说明：

解决1（Python）

#! /usr/bin/env python
#coding:utf-8

#以全局变量X定义节点集合，即类似{'A':'A','B':'B','C':'C','D':'D'},
#如果A、B两点联通，则会更改为{'A':'B','B':'B",...},即任何两点联通之后，两点的值value将相同。

X = dict()      

#各点的初始等级均为0,如果被做为连接的的末端，则增加1

R = dict()

#设置X R的初始值

def make_set(point):
    X[point] = point
    R[point] = 0

#节点的联通分量

def find(point):
    if X[point] != point:
        X[point] = find(X[point])
    return X[point]

#连接两个分量（节点）

def merge(point1,point2):
    r1 = find(point1)
    r2 = find(point2)
    if r1 != r2:
        if R[r1] > R[r2]:
            X[r2] = r1
        else:
            X[r1] = r2
            if R[r1] == R[r2]: R[r2] += 1

#KRUSKAL算法实现

def kruskal(graph):
    for vertice in graph['vertices']:
        make_set(vertice)

    minu_tree = set()
    
    edges = list(graph['edges'])
    edges.sort()                    #按照边长从小到达排序
    for edge in edges:
        weight, vertice1, vertice2 = edge
        if find(vertice1) != find(vertice2):
            merge(vertice1, vertice2)
            minu_tree.add(edge)
    return minu_tree


if __name__=="__main__":

    graph = {
        'vertices': ['A', 'B', 'C', 'D', 'E', 'F'],
        'edges': set([
            (1, 'A', 'B'),
            (5, 'A', 'C'),
            (3, 'A', 'D'),
            (4, 'B', 'C'),
            (2, 'B', 'D'),
            (1, 'C', 'D'),
            ])
        }

    result = kruskal(graph)
    print result

解决2（Python）

#! /usr/bin/env python
#coding:utf-8

class UnionFind:

    def __init__(self):
        
        self.weights = {}
        self.parents = {}

    def __getitem__(self, object):
        
        if object not in self.parents:
            self.parents[object] = object
            self.weights[object] = 1
            return object


        path = [object]
        root = self.parents[object]
        while root != path[-1]:
            path.append(root)
            root = self.parents[root]

        for ancestor in path:
            self.parents[ancestor] = root
        return root
        
    def __iter__(self):
        
        return iter(self.parents)

    def union(self, *objects):
       
        roots = [self[x] for x in objects]
        heaviest = max([(self.weights[r],r) for r in roots])[1]
        for r in roots:
            if r != heaviest:
                self.weights[heaviest] += self.weights[r]
                self.parents[r] = heaviest


#判断是否是无向图
def isUndirected(G):
    
    for v in G:
        if v in G[v]:
            return False
        for w in G[v]:
            if v not in G[w]:
                return False
    return True


#最小生成树

import unittest

def MinimumSpanningTree(G):
    
    if not isUndirected(G):
        raise ValueError("MinimumSpanningTree: input is not undirected")
    for u in G:
        for v in G[u]:
            if G[u][v] != G[v][u]:
                raise ValueError("MinimumSpanningTree: asymmetric weights")

    
    subtrees = UnionFind()
    tree = []
    for W,u,v in sorted((G[u][v],u,v) for u in G for v in G[u]):
        if subtrees[u] != subtrees[v]:
            tree.append((u,v))
            subtrees.union(u,v)
    return tree        


#单元测试

class MSTTest(unittest.TestCase):
    def testMST(self):
        
        G = {0:{1:11,2:13,3:12},1:{0:11,3:14},2:{0:13,3:10},3:{0:12,1:14,2:10}}
        T = [(2,3),(0,1),(0,3)]
        for e,f in zip(MinimumSpanningTree(G),T):
            self.assertEqual(min(e),min(f))
            self.assertEqual(max(e),max(f))

if __name__ == "__main__":
    unittest.main()   

参考资料

1.《算法基础》(GILLES Brassard,Paul Bratley)

2.http://www.ics.uci.edu/~eppstein/PADS/

更多相关代码，请访问：https://github.com/qiwsir/algorithm