2012-03-02 58 views
4

我正試圖解決某個網站的編程挑戰問題,我不想在此提及。優化python代碼,減少分配/釋放時間

的問題是如下:

有一個方形板具有N * N個點。昆蟲從特定點開始(x1,y1)。如果| x1-x2 | + | y1-y2 |,它可以跳到任何點(x2,y2) < = S。另外,有些點含有昆蟲不能跳躍的水。 M跳可以採用多少種不同的路徑?請注意,它可以保持在同一點跳躍到位。

的整數NSM給出,所以是初始板配置。

我很肯定我的解決方案是正確的,可以通過歸納法證明。我將棋盤轉換成一個圖形(鄰接列表),並在兩個點之間形成有效的昆蟲跳躍。那麼它只需要重複M次並更新路徑計數。

我主要關心的是代碼需要進行優化,以便它可以在多個測試用例上工作而不會分配/釋放太多時間,從而減慢運行時間。如果任何人都可以在算法本身中提出優化建議,那將會很棒。

謝謝!

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 

回答

1

這是什麼樣的事情,numpy是好的,雖然你可能能夠通過使用arraystruct模塊來管理它的這個特定使用案例。