kruskal(克鲁斯卡尔)的思路很直观,边按权值从小到大排序,然后从小到大选不会构成回路的边,构成生成树。(选两点不在同一个连通分量里面的边)
构建并查集,用并查集判断是否构成回路(是否在同一个分量里面)(两个连通分量如果根结点相同,两点连接就会构成回路)
python代码:
def find(x, pres):
"""
查找x的最上级(首级)
:param x: 要查找的数
:param pres: 每个元素的首级
:return: 根结点(元素的首领结点)
"""
root, p = x, x # root:根节点, p:指针
# 找根节点
while root != pres[root]:
root = pres[root]
# 路径压缩,把每个经过的结点的上一级设为root(直接设为首级)
while p != pres[p]:
p, pres[p] = pres[p], root
return root
def join(x, y, pres, ranks):
"""
合并两个元素(合并两个集合)
:param x: 第一个元素
:param y: 第二个元素
:param pres: 每个元素的上一级
:param ranks: 每个元素作为根节点时的秩(树的深度)
:return: None
"""
h1, h2 = find(x, pres), find(y, pres)
# 当两个元素不是同一组的时候才合并
# 按秩合并
if h1 != h2:
if ranks[h1] < ranks[h2]:
pres[h1] = h2
else:
pres[h2] = h1
if ranks[h1] == ranks[h2]:
ranks[h1] += 1
def kruskal(n, edges):
"""
kruskal算法
:param n: 结点数
:param edges: 带权边集
:return: 构成最小生成树的边集
"""
# 初始化:pres一开始设置每个元素的上一级是自己,ranks一开始设置每个元素的秩为0
pres, ranks = [e for e in range(n)], [0] * n
# 边从大到小排序
edges = sorted(edges, key=lambda x: x[-1])
mst_edges, num = [], 0
for edge in edges:
if find(edge[0], pres) != find(edge[1], pres):
mst_edges.append(edge)
join(edge[0], edge[1], pres, ranks)
num += 1
else:
continue
if num == n:
break
return mst_edges
# 数据 采用mst图
edges = [
[0, 1, 6],
[0, 2, 1],
[0, 3, 5],
[2, 1, 5],
[2, 3, 5],
[2, 4, 5],
[2, 5, 4],
[1, 4, 3],
[4, 5, 6],
[5, 3, 2]
]
# 结点数
n = 6
mst_edges = kruskal(n, edges)
print('edges:')
for e in mst_edges:
print(e)
print('Total cost of MST:', sum([e[-1] for e in mst_edges]))
print('Maximum cost of MST:', max([e[-1] for e in mst_edges]))
# std print
#
# edges:
# [0, 2, 1]
# [5, 3, 2]
# [1, 4, 3]
# [2, 5, 4]
# [2, 1, 5]
# Total cost of MST: 15
# Maximum cost of MST: 5