In airport of Bytetown, there are two long queues waiting for security check. Checking a person needs one minute, and two queues can be checked at the same time.
Picture from Wikimedia CommonsTwo teams
A and B are going to travel by plane. Each team has n players, ranked from 1 to n according to their average performance. No two players in the same team share the same rank. Team A is waiting in queue 1 while team B is waiting in queue 2. Nobody else is waiting for security check.
Little Q is the policeman who manages two queues. Every time he can check one person from one queue, or check one each person from both queues at the same time. He can‘t change the order of the queue, because that will make someone unhappy. Besides, if two players Ai and Bj are being checked at the same time, satisfying |Ai?Bj|≤k, they will make a lot of noise because their rank are almost the same. Little Q should never let that happen.
Please write a program to help Little Q find the best way costing the minimum time.For each test case, print a single line containing an integer, denoting the minimum time to check all people.
7
Time 2 : Check A_2.
Time 3 : Check A_3.
Time 4 : Check A_4 and B_1.
Time 5 : Check B_2.
Time 6 : Check B_3.
Time 7 : Check B_4.
官方题解：
看题解看了好久，才明白，还是太菜了e_e.
当|ai-bj|<=k时直接转移，当|ai-bj|>k时，可以找到ai的左边第一个at,使得|at-b(j-i+t)|<=k,易知,
t<=k<=i-1这一段,有|ak-b(j-i+k)|>k;所以可以把这一段的转移压缩到只做一次，大大缩短的状态转移所需要的时间
