最近在做項目的過程中遇到了這樣的問題：在有15個節點的無向有環圖中需要求出任意 a，b 兩點間的最短距離路徑。
我的做法是先將圖轉換爲 鄰接矩陣 的形式存儲呈二維數組相連節點爲0不相連爲-1
[[ 0 0 0 0 -1 -1 -1 -1 -1 -1 -1 -1 -1 -1 -1]
[ 0 0 -1 -1 0 -1 -1 -1 -1 -1 -1 -1 -1 -1 -1]
[ 0 -1 0 -1 0 0 -1 -1 -1 -1 -1 -1 -1 -1 -1]
[ 0 -1 -1 0 -1 0 -1 -1 -1 0 -1 -1 0 -1 -1]
[-1 0 0 -1 0 -1 0 -1 -1 -1 -1 -1 -1 -1 -1]
[-1 -1 0 0 -1 0 0 0 0 -1 -1 -1 -1 -1 -1]
[-1 -1 -1 -1 0 0 0 -1 -1 -1 -1 -1 -1 -1 -1]
[-1 -1 -1 -1 -1 0 -1 0 -1 -1 0 -1 -1 -1 -1]
[-1 -1 -1 -1 -1 0 -1 -1 0 0 -1 -1 -1 -1 0]
[-1 -1 -1 0 -1 -1 -1 -1 0 0 -1 -1 -1 0 -1]
[-1 -1 -1 -1 -1 -1 -1 0 -1 -1 0 -1 -1 0 -1]
[-1 -1 -1 -1 -1 -1 -1 -1 -1 -1 -1 1 0 0 0]
[-1 -1 -1 0 -1 -1 -1 -1 -1 -1 -1 0 1 -1 -1]
[-1 -1 -1 -1 -1 -1 -1 -1 -1 0 0 0 -1 0 -1]
[-1 -1 -1 -1 -1 -1 -1 -1 0 -1 -1 0 -1 -1 0]]
然後根據利用遞歸的思想，將問題簡化爲判斷與當前節點相連的其他節點是否存在終點b
所以可根據以下方法求解。
def calculate_short_jump(self, a):
"""
計算無向有環圖中兩個節點間的最短跳數
:param a:
:return:
"""
self.min_nodes = []
self.graph1 = self.get_state(self.cur_state)
# print(self.graph1)
actions = self.get_next([a[0]], a[0])
self.fun(actions, [a[0]], a[1])
res = self.min_nodes[0]
for x in self.min_nodes:
if len(res) > len(x):
res = x
return res
def get_next(self, cur, s):
res = []
for i, x in enumerate(self.graph1[s]):
if x == 0 and i not in cur:
res.append(i)
return res
def fun(self, action_space, cur, s):
if not action_space:
return False
for x in action_space:
if x in cur:
continue
if self.graph1[x][s] == 0:
cur.append(x)
cur.append(s)
stack = cur[:]
self.min_nodes.append(stack)
cur.pop()
cur.pop()
else:
cur.append(x)
actions = self.get_next(cur, x)
res = self.fun(actions, cur, s)
cur.pop()