滴滴：笔试题（20190919）

垃圾分类

利用下三角矩阵求得空闲区块的数目space和冲突的最大值conflict。如果space>=conflict则约束条件不影响分成两个车厢；否则根据space值返回。

def garbage(values, n):
    if not values:
        return n // 2 * 2
    conflict, space = 0, 0
    for nu in range(n):
        current = sum(values[nu][:nu])
        if current == 0:
            space += 1
        if current > conflict:
            conflict = current
    if space >= conflict:
        return n // 2 * 2
    else:
        return space * 2


if __name__ == '__main__':
    n, m = list(map(int, input().strip().split()))
    values = [[0 for _ in range(n)] for _ in range(n)]
    for _ in range(m):
        one, two = list(map(int, input().strip().split()))
        if one < two:
            one, two = two, one
        values[one-1][two-1] = 1
    res = garbage(values, n)
    print(res)

'''
5 2
1 4
3 4

5 4
1 2
2 3
3 4
4 5
'''

简单游戏

懒人模式，暴力求解。提交结果超时，通过45%。

def simple_game(seq, n, m):
    if not seq:
        return 0
    if n <= m:
        return sum(seq)
    res = sum(seq)
    for cursor in range(m, n):
        for si in range(n-cursor+1):
            cur = sum(seq[si:si+cursor])
            # print(cursor, si, seq[si:si + cursor], cur)
            if cur < res:
                res = cur
    return res


if __name__ == '__main__':
    n, m = list(map(int, input().strip().split()))
    seq = list(map(int, input().strip().split()))
    res = simple_game(seq, n, m)
    print(res)

'''
5 3
1 2 3 4 5

5 3
-2 1 -1 -1 -1
'''

（最近更新：2019年09月19日）