Question

我正在尝试从某个网站解决编程挑战问题，我不想在这里提及。

问题如下：

有一个N * N点的方板。昆虫在特定点（x1，y1）开始。如果| x1-x2 | + | y1-y2 |，它可以跳转到任何点（x2，y2）＆lt; = S.此外，一些点含有昆虫不能跳跃的水。 M跳可以采用多少条不同的路径？请注意，它可以通过跳跃保持在同一点。

给出了整数N，S，M，初始板配置也是如此。

我很确定我的解决方案是正确的，可以通过归纳证明。我将电路板转换为图形（邻接列表），其中点之间的边缘使得有效的昆虫跳跃。然后它只需要迭代M次并更新路径计数。

我主要担心的是代码需要进行优化，以便它可以在多个测试用例上运行，而无需分配/解除分配太多次，这会减慢运行时间。如果任何人都可以在算法本身内建议优化，也会很棒。

谢谢！

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

尽管您可以使用numpy和array模块为这个特定用例管理它，但这是struct有用的东西。

优化python代码，减少分配/释放时间

1 个答案: