优化python代码，减少分配/释放时间

Question

我试图解决一个编程挑战问题，从某个网站，我不想在这里提及。在

问题如下：

There is an square board with N*N points. An insect starts out at a particular point (x1,y1). It can jump to any point (x2,y2) if |x1-x2|+|y1-y2| <= S. Also, some points contain water on which the insect cannot jump. How many different paths can it take in M jumps? Note that it can remain at the same point by jumping in place.

给出了整数N、S、M以及初始板配置。在

我很确定我的解决方案是正确的，并且可以通过归纳法加以证明。我把棋盘转换成一个图（邻接列表），点之间有边，可以使昆虫跳跃。那么这只是一个迭代M次和更新路径计数的问题。在

我主要关心的是代码需要进行优化，使其能够在多个测试用例中工作，而不会分配/释放太多时间，这会降低运行时间。如果有人能在算法本身中提出优化建议，那就太好了。在

谢谢！在

import sys

#The board on which Jumping insect moves.
#The max size in any test case is 200 * 200
board = [['_']*200 for j in xrange(200)]

#Graph in the form of an adjancency list created from the board
G = [list() for i in xrange(200*200)]


def paths(N,M,S):
    '''Calculates the total number of paths insect takes
    The board size is N*N, Length of paths: M,
    Insect can jusp from square u to square v if ||u-v|| <=S
    Here ||u-v|| refers to the 1 norm'''

    # Totals paths are modulo 1000000007
    MOD = 1000000007

    # Clearing adjacency list for this testcase
    for i in xrange(N*N): del(G[i][:])

    s = -1 #Starting point s

    #Creating G adjacency list 
    # Point 'L' represents starting point
    # Point 'P' cannot be accessed by the insect
    for u in xrange(N*N):
        x1, y1 = u/N, u%N
        if board[x1][y1] == 'L': s = u
        elif board[x1][y1] == 'P': continue
        for j in xrange(S+1):
            for k in xrange(S+1-j):
                x2, y2 = x1+j, y1+k
                if x2 < N and y2 < N and not board[x2][y2] == 'P':
                    v = x2*N+y2
                    G[u].append(v)
                    if not u == v: G[v].append(u)
                if j > 0 and k > 0:
                    x2, y2 = x1+j, y1-k
                    if x2 < N and y2 >= 0 and not board[x2][y2] == 'P':
                        v = x2*N+y2
                        G[u].append(v)
                        G[v].append(u)                

    # P stores path counts
    P = [[0 for i in xrange(N*N)] for j in xrange(2)]
    # Setting count for starting position to 1
    P[0][s] = 1

    # Using shifter to toggle between prev and curr paths
    shifter, prev, curr = 0, 0, 0

    # Calculating paths
    for i in xrange(M):
        prev, curr = shifter %2, (shifter+1)%2

        #Clearing Path counts on curr
        for i in xrange(N*N): P[curr][i] = 0 
        for u in xrange(N*N):
            if P[prev][u] == 0: continue
            for v in G[u]:
                P[curr][v] = (P[curr][v]+P[prev][u]) % MOD
        shifter = (shifter+1)%2
    return (sum(P[curr])%MOD)

#Number of testcases
num = int(sys.stdin.readline().split()[0])

results = []

# Reading in test cases
for i in xrange(num):
    N, M, S = [int(j) for j in sys.stdin.readline().split()]
    for j in xrange(N):
        board[j][:N] = list(sys.stdin.readline().split()[0])
    results.append(paths(N,M,S))

for result in results:
    print result

Answer 1

这正是numpy的优点，尽管您可以通过使用array和struct模块来管理这个特定用例。在