AGC049B

B - Flip Digits

from AGC049 AGC049B

• Thoughts.

  • The operation to be performed is either 01 to 10 or 11 to 00

  • Thinking from behind

    • 
    • In case 1, trailing 1s are bears.
      • Because it can’t disappear.
    • Can I turn it off in case 2?
    • DP by the number of times it can be deleted?
      • N=10^5, so it’s hard when you can delete about 10
    • In case 2, can I turn it off? →No, not good.
      • S: 110011
      • T: 001100
      • At this time, if the back is erased first, it becomes unattainable.
      • Do you want to do it before?

  • I’ve been doing this for a while.

    • 
      • Greed Law Proof Patterns
    • If the pattern to be deleted is in the optimal solution, it contradicts the assumption that it is the optimal solution because a better solution can be created by exchanging the places to be deleted, and therefore there is no pattern to be deleted
    • AC

  • Tweet:

    • The operation is to move one 1 to the left or erase 2 pieces.
    • When we focus on the leftmost 1 and S is to the left, we add the cost of erasing the 1 to the answer to make the 1 we focus on two ahead, because if there is a solution, that 1 must disappear.
    • In the reverse case, the first 1 is not eliminated in the optimal solution because eliminating the first 1 is a loss.
      • Add the cost of moving the first 1 in S to the first 1 in T to the answer and focus on the next 1, respectively.
    • When you reach the end, it is the best solution and you return the answer.
    • If an inconsistency occurs, such as a missing S, we know that a solution does not exist.
    • My implementation enumerates where one stands first, but this may be solved without doing so.
      • The code is easy to understand because “look out for the next one” is a subscript increment. python

def solve(N, S, T):
    spos = []
    tpos = []
    diff = 0
    for i in range(N):
        if S[i] == 1:
            spos.append(i)
            diff += 1
        if T[i] == 1:
            tpos.append(i)
            diff -= 1

    if diff % 2 == 1:
        return -1
    if diff < 0:
        return -1

    tpos += [N] * N
    spos.append(-1)
    i = 0
    j = 0
    ret = 0
    while i < len(spos) - 1:
        if spos[i] < tpos[j]:
            # spos_i should be deleted
            next = spos[i + 1]
            if next == -1:
                return -1
            ret += (next - spos[i])
            i += 2
            continue

        ret += (spos[i] - tpos[j])
        i += 1
        j += 1
    if j != len(tpos) - N:
        return -1
    return ret

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