shortest path 实现

使用python内置函数直接实现Dij算法，每次用全部已包含集中的点进行计算：

def Dij(start):list_a = [1,2,3,4,5,6,7,8,9,10] ##所有点的集合list_s = [start]  ##最短路径已经确定的点的集合sdist = {} ##最后的结果，存为一个dictionary，key是点的数字，value是从start出发最短的距离dist[start] = 0 while True:if len(list_s) == 10: return disttemp = {} ##临时字典，记录所有从s出发右边可以用一条边连接的点for i in list_s:for j in list(set(list_a)-set(list_s)):if map[i][j] >= 0:temp[(dist[i]+map[i][j])] = [i,j] ##记录点的临时最短距离，为s中的点的最短距离加上该边的距离dist[temp[min(temp)][1]] = min(temp) ##挑选其中临时最短距离最小的一个点加入s，其临时最短距离就是其真实的最短距离patch[temp[min(temp)][1]] = temp[min(temp)][0] ##记录最短路径具体经过的点list_s.append(temp[min(temp)][1]) ##挑选其中临时最短距离最小的一个点加入sif __name__ == '__main__':map=[[],[[],-1,-1,-1,-1,-1,-1,-1,-1,-1,-1],[[],-1,-1,-1,-1,-1,-1,-1,-1,-1,-1],[[],-1,-1,-1,-1,-1,-1,-1,-1,-1,-1],[[],-1,-1,-1,-1,-1,-1,-1,-1,-1,-1],[[],-1,-1,-1,-1,-1,-1,-1,-1,-1,-1],[[],-1,-1,-1,-1,-1,-1,-1,-1,-1,-1],[[],-1,-1,-1,-1,-1,-1,-1,-1,-1,-1],[[],-1,-1,-1,-1,-1,-1,-1,-1,-1,-1],[[],-1,-1,-1,-1,-1,-1,-1,-1,-1,-1],[[],-1,-1,-1,-1,-1,-1,-1,-1,-1,-1],]map[1][2] = 40map[1][3] = 8map[1][4] = 10map[2][7] = 10map[2][5] = 6map[3][2] = 4map[3][4] = 12map[3][6] = 2map[4][6] = 1map[5][3] = 2map[5][6] = 2map[5][7] = 4map[6][8] = 4map[6][9] = 3map[7][8] = 20map[7][10] = 1map[8][5] = 0map[8][10] = 20map[9][4] = 6map[9][8] = 10map[9][10] = 2patch={}Dij(1)

使用data_structure课中的，每次用结点进行搜索，并不断更新dist的步骤：

Python实现代码：

def Dij(start):list_a = [1,2,3,4,5,6,7,8,9,10]list_s = [start]  dist = [[],100,100,100,100,100,100,100,100,100,100]dist[start] = 0 ##初始化最短距离，起点设为0，和起点有边连接的设为该边数值，其余点设为无穷大（此处为100）for j1 in list_a: if map[start][j1] >= 0:dist[j1] = map[start][j1]while True:if len(list_s) == 10: ##终止条件return disttemp ={}for j2 in list(set(list_a)-set(list_s)): ##未收录在s集合中的点中dist最小的点Vtemp[dist[j2]] = j2V = temp[min(temp)]list_s.append(V) ##将V加入s中for j3 in list(set(list_a)-set(list_s)): ##更新所有未收录在s集中的点的distif map[V][j3] >= 0:if dist[V] + map[V][j3] < dist[j3]:dist[j3] = dist[V] + map[V][j3]path[j3] = V

现在考虑解决任意起点终点的最短路径问题，用floyd算法：

每次迭代计算只经过编号小于k的定点的路径

其Python的代码实现为：

def floyd():## 初始化时没有边直接相连的点之间最短距离需要设置成无穷大,此处为100P = mapD = mapfor i1 in range(1,11):for j1 in range(1,11):if i1 == j1:D[i1][j1] = 0if map[i1][j1] == -1 :D[i1][j1] = 100for k in range(1,11):for i2 in range(1,11):for j2 in range(1,11):if D[i2][k]+D[k][j2] < D[i2][j2]:D[i2][j2] = D[i2][k]+D[k][j2]return Dif __name__ == '__main__':map=[[],[[],-1,-1,-1,-1,-1,-1,-1,-1,-1,-1],[[],-1,-1,-1,-1,-1,-1,-1,-1,-1,-1],[[],-1,-1,-1,-1,-1,-1,-1,-1,-1,-1],[[],-1,-1,-1,-1,-1,-1,-1,-1,-1,-1],[[],-1,-1,-1,-1,-1,-1,-1,-1,-1,-1],[[],-1,-1,-1,-1,-1,-1,-1,-1,-1,-1],[[],-1,-1,-1,-1,-1,-1,-1,-1,-1,-1],[[],-1,-1,-1,-1,-1,-1,-1,-1,-1,-1],[[],-1,-1,-1,-1,-1,-1,-1,-1,-1,-1],[[],-1,-1,-1,-1,-1,-1,-1,-1,-1,-1],]map[1][2] = 40map[1][3] = 8map[1][4] = 10map[2][7] = 10map[2][5] = 6map[3][2] = 4map[3][4] = 12map[3][6] = 2map[4][6] = 1map[5][3] = 2map[5][6] = 2map[5][7] = 4map[6][8] = 4map[6][9] = 3map[7][8] = 20map[7][10] = 1map[8][5] = 0map[8][10] = 20map[9][4] = 6map[9][8] = 10map[9][10] = 2result = floyd()

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