一．题目要求

参考下图完成游戏地图中从起点到目标点的最短路径寻找问题。

二．设计思路

先对游戏地图做了几个设定，以矩阵来模拟游戏地图。将可行的区域位置赋值0，障碍区赋值为inf。考虑到地图大小，将起始点和终点区域赋值99。

从Start点A开始向外层扩展，每扩展一层pathlen加一。List Q存储当前需要扩展的点，list P 存储当前扩展层。当扩展到End点B时扩展结束，路径可规划。当Q为空时，本次层扩展结束，检查P，若P非空，从P层向外扩展，若P为空，则End点B无法到达。

寻找最短路径时，从End点B开始，寻找当前点附近8个点的标记中比当前点标记小的点，直到标记为1为止。

三．程序主体

# -*-coding:gbk -*-
from numpy import *
dirs = [(1,1),(1,0),(1,-1),(0,-1),(-1,-1),(-1,0),(-1,1),(0,1)]  # 四邻位置：从右下角开始顺时针得到，是按坐标差得到的
def find_path(oldmap,A,B):
    oldmap[A[0], A[1]] = 99
    oldmap[B[0], B[1]] = 99
    [a,b]=oldmap.shape
    pathmap=oldmap.copy()
    Q=[]#存储扩展节点
    P=[]#往外一层
    pathlen=1
    if A==B:
        print('start point is equal to end point')
        return True
    current=A
    while (True):
        for i in range(8):
            neighbor=[current[0]+dirs[i][0], current[1]+dirs[i][1]]
            if neighbor==B:
                print('the way is found')######################wrong
                print('中间过程')
                print(oldmap)
                find_way(oldmap,pathmap,A,B,a,b)#####调用路径函数
                return True
            if (neighbor[0]>=0 and neighbor[1]>=0 and neighbor[0]<a and neighbor[1]<b and oldmap[neighbor[0],neighbor[1]]==0):
                P.append(neighbor)
                oldmap[neighbor[0],neighbor[1]]=pathlen

        if Q==[]:
            if P ==[]:
                print(oldmap)  ##############
                print('No path')
                return False
            else:
                Q.extend(P)
                P=[]
                pathlen += 1

        else:
            current=Q.pop()

###################寻找最短路径
def find_way(oldmap,pathmap,A,B,a,b):
    currentpos=B
    while (oldmap[currentpos[0],currentpos[1]]!=1):
        for i in range(8):
            neighborpos=[currentpos[0]+dirs[i][0], currentpos[1]+dirs[i][1]]
            if (neighborpos[0] >= 0 and neighborpos[1] >= 0 and neighborpos[0] < a and neighborpos[1] < b and oldmap[neighborpos[0],neighborpos[1]]!=0):
                if oldmap[neighborpos[0],neighborpos[1]]<oldmap[currentpos[0],currentpos[1]]:
                   pathmap[neighborpos[0],neighborpos[1]]=oldmap[neighborpos[0],neighborpos[1]]
                   currentpos=neighborpos
                   break
    print('the way:')
    print(pathmap)

四．主函数

def main():
    map =mat([[0,    0,  0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0],
              [0,    0,  0, 0, inf,inf, 0, 0, 0, 0, 0, 0, 0],
              [0,    0,  0, 0, 0,inf, 0, 0, 0, 0, 0, 0, 0],
              [inf,inf,inf, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0],
              [0,    0,inf, 0, 0, 0, 0, 0, 0, 0, 0, 0, inf],
              [0,    0,inf, 0, 0, 0, 0, 0, 0, 0, 0, 0, inf],
              [0,    0,inf, 0, 0, 0, 0, 0, 0, 0, 0, 0,inf],
              [0,    0,  0, 0, 0, 0, 0, 0, 0, 0, 0, 0, inf],
              [0,    0,  0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0],])
    print('最初地图')
    print(map)
    print('**********************************')
    A = [5, 0]
    # B=[5,0]
    B = [3, 12]
    find_path(map,A, B)


if __name__=='__main__':
    main()

五．运行结果

六．结果分析

由中间过程对应的矩阵可知，共经历了12次向外层扩展，第12次扩展即可将目标点包含进去。最短路径如the way对应的矩阵所示，是通过一种类似梯度下降的方法得到的。