Question

现在比赛结束了，所以我想问我的算法在我的代码上失败了。

This is the problem. If anyone interested, you can see it at here

def solve():
    S = int(input())
    D, A, B, M, N = [], [], [], [], []

    for i in range(S):
        d, a, b = [int(c) for c in input().split(" ")]
        D.append(d); A.append(a); B.append(b)
        M.append(d+a); N.append(d-b)

    straightMstart, straightNstart = [0]*(S), [0]*(S)
    crossedMstart, crossedNstart = [0]*(S), [0]*(S) # cross over
    for i in range(1, S):
        if M[i] == M[i-1]:
            straightMstart[i] = straightMstart[i-1]
            crossedNstart[i] = crossedNstart[i-1]
        else:
            straightMstart[i] = i

        if N[i] == N[i-1]:
            straightNstart[i] = straightNstart[i-1]
            crossedMstart[i] = crossedMstart[i-1]
        else:
            straightNstart[i] = i

        if M[i] != M[i-1]:
            crossedNstart[i] = straightNstart[i-1]
        if N[i] != N[i-1]:
            crossedMstart[i] = straightMstart[i-1]

    maxlen = 1
    maxlensubsets = 1
    for i in range(1, S):
        thislen = i - min(crossedMstart[i], crossedNstart[i]) + 1
        if maxlen < thislen:
            maxlen = thislen
            maxlensubsets = 1
        elif maxlen == thislen:
            maxlensubsets += 1

    # print(crossedNstart)
    # print(crossedMstart)
    return "%d %d" % (maxlen, maxlensubsets)

testcase = int(input())
for tc in range(1, testcase+1):
    print("Case %d: %s" % (tc, solve()))

我使用交叉最大长度来找到最大的集合大小（对于M和Ns）

我将举出以下示例，以便更容易理解我的逻辑：

# Let's suppose that the M and N is:
M (=D[i]+A[i]) = [  9,  9, 18, 22, 22]
N (=D[i]-B[i]) = [-10, -5,  7, -1, -1]

# Straight Start means starting index for same value.
# M=9 starts from index 0, M=18 starts from index 2, M=22 starts from index 3
straightMstart = [0, 0, 2, 3, 3]
# Same logic applied to straightNstart
straightNstart = [0, 1, 2, 3, 3]

# Crossed Start means cross-starting index of opponent array.    
# For crossedMstart, you start from N[i] and climb N then cross to climb M
# The reason why I swapped order of cNs and cMs is that both arrays are based on opponent arrays
crossedNstart = [0, 0, 1, 2, 2]
crossedMstart = [0, 0, 0, 2, 2]

我真的很困惑，我真的不明白我的错是什么意思。请帮我纠正我的逻辑。

Answer 1

您的算法有多种失败方式。最简单的是在任何输入中输入+-----------+-----------+ | 1 | 2 | +-----------+-----------+ | … | … | +-----------+-----------+ | 19 | 20 | +-----------+-----------+。然后，我确实发现string存在问题，如果straightNstart[i]或N[i]匹配M[i]或N[i-1]，则当前迭代始终为零。这会将M[i-1]和crossedMstart[i]值设置为crossedNstart[i]。另一个问题是当0忽略两个循环时。

第一个问题很容易解决，可能会解决d，a，b，testcase的问题，其中d，a，b，testcase应该只引发一个TypeError：

S <= 1

对于另一个问题，可以迭代两次，以便S = input("[description]") if S.isdigit(): S = int(S) if S <= 1: raise ValueError("Value must be greater then 1") else: raise TypeError("Value must be numeric")，straightNstart[i]分配对结果有任何影响（至少当M [i]或N [i]是不等于M [i-1]或N [i-1]）。

Answer 2

我发现我的算法不处理多个交叉。

我不知道我的算法在Google Code Jam 2018 1B-B的代码中出了什么问题

2 个答案: