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A算法-八数码问题.py

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+# 目标序列end = [1,2,3,8,0,4,7,6,5]
+# end = 1 2 3
+#       8 0 4
+#       7 6 5
+import sys
+
+
+class Statusobj:
+    def __init__(self):
+        # 状态序列
+        self.array = []
+        # 该状态下的深度
+        self.deep = 0
+        # 该状态的f(n) = g(n) + h(n)评估函数值
+        self.Fn = 0
+        # 该状态由上一个状态经何种操作得到，避免出现死循环
+        # 1:up  2:down  3:left  4:right
+        self.preoperate = 0
+        self.Father = Statusobj
+
+
+# 判断八数码问题是否有解，在于若两个状态的逆序奇偶性 相同，则可相互到达，否则不可相互到达
+def getreversenum(array):  # 获取当前状态的逆序数和
+    count = 0
+    for i in range(len(array)):
+        for j in range(i+1, len(array)):
+            if array[j] < array[i]:
+                count += 1
+    return count
+
+
+def countnotinplace(array, end):  # 统计当前状态和目标状态相比较的不在位将牌数
+    count = 0
+    for i in range(len(array)):
+        if array[i] != end[i] and array[i] != 0:
+            count += 1
+    return count
+
+
+def selectoperate(i, preoperate):  # 通过当前状态下的序列下标和上一次操作符得到可选择的操作
+    selected = []
+    if 3 <= i <= 8 and preoperate != 2:
+        selected.append(1)
+    if 0 <= i <= 5 and preoperate != 1:
+        selected.append(2)
+    if i in [1, 2, 4, 5, 7, 8]:
+        if preoperate != 4:
+            selected.append(3)
+    if i in [0, 1, 3, 4, 6, 7]:
+        if preoperate != 3:
+            selected.append(4)
+    return selected
+
+
+def op(i, x):  # 通过操作符得到新下标
+    if i == 1:  # up
+        return x-3
+    if i == 2:  # down
+        return x+3
+    if i == 3:  # left
+        return x-1
+    if i == 4:  # right
+        return x+1
+
+
+def newarray(operate, array):  # 通过操作符拓展的新节点
+    subscript = array.index(0)
+    newindex = op(operate, subscript)
+    temp = array[subscript]
+    array[subscript] = array[newindex]
+    array[newindex] = temp
+    return array
+
+
+def printarray(array):
+    print(f"{array[:3]}\n{array[3:6]}\n{array[6:]}\n", end='')
+
+
+def main():
+    openlist = []  # open表，保存待扩展实例对象
+    closelist = []  # close表
+    endarray = [1, 2, 3, 8, 0, 4, 7, 6, 5]
+    initobj = Statusobj()
+    initobj.array = list(map(int, input("请输入原始状态的八数码序列：").split()))
+    initobj.deep = 0  # 初始化执行深度
+    initobj.Fn = initobj.deep + countnotinplace(initobj.array, endarray)
+    openlist.append(initobj)
+    array_reverse = getreversenum(initobj.array) - initobj.array.index(0)
+    end_reverse = getreversenum(endarray) - endarray.index(0)
+    if array_reverse % 2 == end_reverse % 2:
+        while True:
+            if openlist[0].array == endarray: # 达到目标状态
+                end_list = []
+                end_list.append(openlist[0])
+                deep = openlist[0].deep
+                closelist.append(openlist[0])
+                del openlist[0]
+                father = closelist[-1].Father
+                while father.deep >= 1:
+                    end_list.append(father)
+                    father = father.Father
+                end_list.append(initobj)
+                print("目标状态：")
+                printarray(endarray)
+                print("【变换成功,共需要" + str(deep) + "次变换】")
+                d = 0
+                for it in reversed(end_list):
+                    print("当前深度为：{}".format(d))
+                    d += 1
+                    print("操作序号为：{}".format(it.preoperate))
+                    printarray(it.array)
+                sys.exit()
+            else:
+                zero_index = openlist[0].array.index(0)
+                temp_list = []
+                operation = selectoperate(zero_index, openlist[0].preoperate)
+                for ope in operation:
+                    copy_array = openlist[0].array.copy()  # .copy()复制序列，必不可少
+                    new_array = newarray(ope, copy_array)
+                    new_obj = Statusobj()
+                    new_obj.array = new_array  # 序列
+                    new_obj.deep = openlist[0].deep + 1  # 深度
+                    new_obj.preoperate = ope  # 上一操作符
+                    new_obj.Father = openlist[0]  # 父亲节点
+                    new_obj.Fn = new_obj.deep + countnotinplace(new_array, endarray)  # 评估函数
+                    temp_list.append(new_obj)
+                closelist.append(openlist[0])
+                del openlist[0]
+                for item in temp_list:
+                    openlist.append(item)
+                openlist.sort(key=lambda x: x.Fn)
+    else:
+        print("此情况下无解！")
+
+
+main()
+# 2 8 3 1 6 4 7 0 5