我想研究一些“最大流量”问题来理解这些算法，但是我对于测试它们的简单设置的概念证明难以实现。
看看这个Project Euler问题：http://projecteuler.net/problem=83
我想要做的是假设每个单元都连接到它的所有相邻单元（“+”模式），然后在每个成对==之间创建一条路径到它们两者之间的最大值即：max(cell1, cell2）
因此，一个简单的[[s 4],[3 t]]矩阵将成为[(s, (0,1), 4), (s, (1,0), 3), ((0,1), t, 3), ((1,0), 4, t)]（node1，node2，cost）+所有路径都朝另一个方向发展。
也许有一种更简单的方式来描述我想做的事情，但我会感激任何帮助。
其他细节：我正在使用Python和NumPy。
答案 0 :(得分：-1)
这是用于计算Project Euler问题83的边缘的ipython笔记本代码。我使用索引代替矩阵中的每个元素，而不是2D坐标。
In [1]:
#load the data
import numpy as np
from StringIO import StringIO
data = StringIO("""131 673 234 103 18
201 96 342 965 150
630 803 746 422 111
537 699 497 121 956
805 732 524 37 331""")
m = np.loadtxt(data, dtype=int)
m
Out[1]:
array([[131, 673, 234, 103, 18],
[201, 96, 342, 965, 150],
[630, 803, 746, 422, 111],
[537, 699, 497, 121, 956],
[805, 732, 524, 37, 331]])
In [2]:
# give every element an index
idx = np.arange(m.size).reshape(m.shape)
idx
Out[2]:
array([[ 0, 1, 2, 3, 4],
[ 5, 6, 7, 8, 9],
[10, 11, 12, 13, 14],
[15, 16, 17, 18, 19],
[20, 21, 22, 23, 24]])
In [3]:
# create edges
left_edges = np.concatenate([idx[:, :-1, None], idx[:, 1:, None], m[:, 1:, None]], axis=2).reshape(-1, 3)
right_edges = np.concatenate([idx[:, 1:, None], idx[:, :-1, None], m[:, :-1, None]], axis=2).reshape(-1, 3)
down_edges = np.concatenate([idx[:-1, :, None], idx[1:, :, None], m[1:, :, None]], axis=2).reshape(-1, 3)
up_edges = np.concatenate([idx[1:, :, None], idx[:-1, :, None], m[:-1, :, None]], axis=2).reshape(-1, 3)
edges = np.vstack((left_edges, right_edges, down_edges, up_edges))
edges
Out[3]:
array([[ 0, 1, 673],
[ 1, 2, 234],
[ 2, 3, 103],
[ 3, 4, 18],
[ 5, 6, 96],
[ 6, 7, 342],
[ 7, 8, 965],
[ 8, 9, 150],
[ 10, 11, 803],
[ 11, 12, 746],
[ 12, 13, 422],
[ 13, 14, 111],
[ 15, 16, 699],
[ 16, 17, 497],
[ 17, 18, 121],
[ 18, 19, 956],
[ 20, 21, 732],
[ 21, 22, 524],
[ 22, 23, 37],
[ 23, 24, 331],
[ 1, 0, 131],
[ 2, 1, 673],
[ 3, 2, 234],
[ 4, 3, 103],
[ 6, 5, 201],
[ 7, 6, 96],
[ 8, 7, 342],
[ 9, 8, 965],
[ 11, 10, 630],
[ 12, 11, 803],
[ 13, 12, 746],
[ 14, 13, 422],
[ 16, 15, 537],
[ 17, 16, 699],
[ 18, 17, 497],
[ 19, 18, 121],
[ 21, 20, 805],
[ 22, 21, 732],
[ 23, 22, 524],
[ 24, 23, 37],
[ 0, 5, 201],
[ 1, 6, 96],
[ 2, 7, 342],
[ 3, 8, 965],
[ 4, 9, 150],
[ 5, 10, 630],
[ 6, 11, 803],
[ 7, 12, 746],
[ 8, 13, 422],
[ 9, 14, 111],
[ 10, 15, 537],
[ 11, 16, 699],
[ 12, 17, 497],
[ 13, 18, 121],
[ 14, 19, 956],
[ 15, 20, 805],
[ 16, 21, 732],
[ 17, 22, 524],
[ 18, 23, 37],
[ 19, 24, 331],
[ 5, 0, 131],
[ 6, 1, 673],
[ 7, 2, 234],
[ 8, 3, 103],
[ 9, 4, 18],
[ 10, 5, 201],
[ 11, 6, 96],
[ 12, 7, 342],
[ 13, 8, 965],
[ 14, 9, 150],
[ 15, 10, 630],
[ 16, 11, 803],
[ 17, 12, 746],
[ 18, 13, 422],
[ 19, 14, 111],
[ 20, 15, 537],
[ 21, 16, 699],
[ 22, 17, 497],
[ 23, 18, 121],
[ 24, 19, 956]])
然后，您可以edges
scipy.sparse.coo_matrix转换为稀疏矩阵