[英]Graph initialisation using dictionaries for floyd algorithm on undirected graphs?
我想知道如何在无向图上实现floyd。 通过此实施,
for k in graph:
for i in graph:
for j in graph:
if dist[i][k] + dist[k][j] < dist[i][j]:
dist[i][j] = dist[i][k] + dist[k][j]
pred[i][j] = pred[k][j]
我能够得到以下的邻接矩阵:
__0 __1 __2 __3 __4
0| 0 28 38 33 63
1| inf 0 10 60 44
2| inf inf 0 50 80
3| inf inf inf 0 30
4| inf inf inf inf 0
这是有向图。 很好,但是我希望邻接矩阵显示无向图。 据我了解,对于无向图,这些路径应沿对角线0镜像,但是我不确定如何做到这一点。
我试过了:
for k in graph:
for i in graph:
for j in graph:
if dist[i][k] + dist[k][j] < dist[i][j]:
dist[i][j] = dist[i][k] + dist[k][j]
dist[j][i] = dist[i][k] + dist[k][j]
pred[i][j] = pred[k][j]
但是我的图形结果是错误的,并且看起来该算法正在经历不需要的边缘以到达所需的顶点。
__0 __1 __2 __3 __4
0| 0 48 38 93 63
1| 48 0 110 60 44
2| 38 110 0 110 80
3| 93 60 110 0 30
4| 63 inf 80 inf 0
{0 : {1:28, 3:33},
1 : {2:10, 4:44},
2 : {3:50},
3 : {4:30},
4 : {}}
编辑:这是完整的代码
import ast
def floyd(graph):
dist = {}
pred = {}
for u in graph:
dist[u] = {}
pred[u] = {}
for v in graph:
dist[u][v] = float('INF')
pred[u][v] = -1
dist[u][u] = 0
for z in graph[u]:
dist[u][z] = graph[u][z]
pred[u][z] = u
for k in graph:
for i in graph:
for j in graph:
if dist[i][k] + dist[k][j] < dist[i][j]:
dist[i][j] = dist[i][k] + dist[k][j]
return dist, pred
graph = {}
graph = ast.literal_eval(open("file2.txt").read())
print graph
dist, pred = floyd(graph)
print " ",
for j in dist: print "__" + str(j),
print "\n"
for i in dist:
print str(i) + "|",
for v in dist: print " %s" % (dist[i][v]),
print "\n"
我认为,只需镜像图形字典中的所有边缘即可解决您的问题,例如:
graph = {0 : {1:28, 3:33},
1 : {2:10, 4:44},
2 : {3:50},
3 : {4:30},
4 : {}}
# Mirror all edges
for u in graph:
for v in graph[u]:
if not u in graph[v]:
graph[v][u] = graph[u][v]
这是将输入设置为无向图的最简单方法(尽管显然,如果您随后编辑/删除边,则需要更加小心)。 当我将其插入您的代码时,我得到:
__0__1__2__3__4
0| 0 28 38 33 63
1| 28 0 10 60 44
2| 38 10 0 50 54
3| 33 60 50 0 30
4| 63 44 54 30 0
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