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How to solve spoj CLEVER?
By rebornplusplus, history, 6 years ago, In English

How to solve spoj's CLEVER ?
( Problem statement given below )

CLEVER — The Clever Typist
no tags 
Time limit:	1s
Source limit:	50000B
Memory limit:	1536MB

Blue Mary is a typist of some secret department.Now she has to type in many passwords in an hour,each of which has a fixed length: 6.Of course,the less times she presses the keyboard,the happier she is.

Unfortunately,the keyboard to type in the password is extraordinary designed to keep secrets.The keyboard has 6 particular keys instead of 10 number keys.To explain the usages of these keys,let's define the 6 position on the screen 1,2,3,4,5,6 from left to right.The keys' usages are shown below:

    Swap0: swap the digit in the cursor position and the digit in position 1.The cursor doesn't move.If the cursor is now in position 1,the digits on the screen won't be changed.
    Swap1: swap the digit in the cursor position and the digit in position 6.The cursor doesn't move.If the cursor is now in position 6,the digits on the screen won't be changed.
    Up: increase the digit in the cursor position by 1.If the digit in the cursor position is 9,no change will happen.
    Down: decrease the digit in the cursor position by 1.If the digit in the cursor position is 0,no change will happen.
    Left: move the cursor one position left.If the cursor is in position 1,no change will happen.
    Right: move the cursor one position right.If the cursor is in position 6,no change will happen.

At start,6 random digits will be given on the screen,and the cursor will in position 1.After some smart presses,she can type in the correct password,at that time the cursor position is unimportant.

Here is an example("()"denotes to the cursor):

key pressed       screen
                  (1)23456
Swap1             (6)23451
Right             6(2)3451
Swap0             2(6)3451
Down              2(5)3451
Right             25(3)451
Up                25(4)451
Right             254(4)51
Down              254(3)51
Right             2543(5)1
Up                2543(6)1
Swap0             6543(2)1

Now Mary wants to know the minimal number of keys she has to press.Can you help her?
Input

The first line contains a single integer t(about 1000).t lines follow,each contains two 6-digit string,which show the digits on the screen at start and the password Mary is to type in,separated by a single space.
Output

t lines,each contains a single integer — the answer.
Example

Sample input:
1
123456 654321

Sample output:
11
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6 years ago, # |
  Vote: I like it +5 Vote: I do not like it

Dude There is no need of pasting statement,if u already wrote link!!

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6 years ago, # |
← Rev. 2   Vote: I like it 0 Vote: I do not like it

Looks like standard BFS.

Let DP[X][V], be the shortest distance to the situation where your cursor is at x (0 to 5), and the number of the screen is v (0 to 1e6 — 1). Then just put pairs {x, v} into the queue, and stop when you reach the wanted state.

If that TLE's you can try BFS from both ends.

EDIT: I missed that there are multiple queries

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    6 years ago, # ^ |
      Vote: I like it +22 Vote: I do not like it

    what is reason behind ur cf handle name.u deosn't seem indian

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    6 years ago, # ^ |
      Vote: I like it +8 Vote: I do not like it

    @mango_lassi, none of them passed TL actually. My implementation might be an issue, though. But I tried to do as much as I could. My code for bfs from both ends: here.

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      6 years ago, # ^ |
      ← Rev. 2   Vote: I like it 0 Vote: I do not like it

      I missed that the problem has queries, sorry.

      That said, I think I have a solution now.

      For every permutation, precalculate the minimum number of steps to swap the numbers around such that if the numbers first were in sorted order, they'd be in the permutation's order afterwards.

      Additionally, you need to store whether you have at least once been in the same node as the last element, and where the cursor ends up in, for reasons explained later.

      You can do the precalculation with BFS from the initial state.

      Now when we have this information, we can for every query, go over all 720 permutations, find out how much we need to add to or subtract from every integer, and take the best of these offers.

      Of course, you still need to make sure that you have at least once been in the same tile as each number you will change. You can make sure that this has happened, by the "visit last" bit being set if the last number needs to be modified, and cursor position being greater than or equal to index i if index i needs to be modified.

      This way the time use should be something like 720*6*2 + 1000*720*6 = ~4e6

      I'm writing the code now, results soon.

      EDIT: turns out you can't just store the position of the cursor, you need to store the highest position the cursor has ever been in.

      AC code

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        6 years ago, # ^ |
        ← Rev. 2   Vote: I like it 0 Vote: I do not like it

        Would you please explain why I need to store the max cursor position? I am not very clear on that. :/

        EDIT: Understood. Thanks a ton!

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