zoukankan      html  css  js  c++  java
  • HDU 6076

    Security Check

    Problem Description
    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 Commons


    Two 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 |AiBj|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.
     
    Input
    The first line of the input contains an integer T(1T15), denoting the number of test cases.
    In each test case, there are 2 integers n,k(1n60000,1k10) in the first line, denoting the number of players in a team and the parameter k.
    In the next line, there are n distinct integers A1,A2,...,An(1Ain), denoting the queue 1 from front to rear.
    Then in the next line, there are n distinct integers B1,B2,...,Bn(1Bin), denoting the queue 2 from front to rear.
     
    Output
    For each test case, print a single line containing an integer, denoting the minimum time to check all people.
     
    Sample Input
    1 4 2 2 3 1 4 1 2 4 3
     
    Sample Output
    7
    Hint
    Time 1 : Check A_1.
    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

  • 相关阅读:
    树状数组
    POJ 1178 -- 国王和骑士
    read
    优先队列
    统计八连块
    1579、Function Run Fun(记忆化搜索)
    5488: 石子归并II (区间DP+环形DP+四边形不等式优化)
    4797: 能量项链(区间DP,环形DP)
    5936 桃子的矩阵快速幂
    Happy Necklace(找规律+矩阵快速幂)
  • 原文地址:https://www.cnblogs.com/siyecaodesushuo/p/7290994.html
Copyright © 2011-2022 走看看