Problem:


Score : 400 points

Problem Statement

Along a road running in an east-west direction, there are A shrines and B temples.
The i-th shrine from the west is located at a distance of s_i meters from the west end of the road, and the i-th temple from the west is located at a distance of t_i meters from the west end of the road.

Answer the following Q queries:

Query i (1 \leq i \leq Q): If we start from a point at a distance of x_i meters from the west end of the road and freely travel along the road, what is the minimum distance that needs to be traveled in order to visit one shrine and one temple? (It is allowed to pass by more shrines and temples than required.)

Constraints

1 \leq A, B \leq 10^5

1 \leq Q \leq 10^5

1 \leq s_1 < s_2 < ... < s_A \leq 10^{10}

1 \leq t_1 < t_2 < ... < t_B \leq 10^{10}

1 \leq x_i \leq 10^{10}

s_1, ..., s_A, t_1, ..., t_B, x_1, ..., x_Q are all different.

All values in input are integers.

Input

Input is given from Standard Input in the following format:

A B Q
s_1
:
s_A
t_1
:
t_B
x_1
:
x_Q

Output

Print Q lines. The i-th line should contain the answer to the i-th query.

Sample Input 1

2 3 4
100
600
400
900
1000
150
2000
899
799

Sample Output 1

350
1400
301
399

There are two shrines and three temples. The shrines are located at distances of 100, 600 meters from the west end of the road, and the temples are located at distances of 400, 900, 1000 meters from the west end of the road.

Query 1: If we start from a point at a distance of 150 meters from the west end of the road, the optimal move is first to walk 50 meters west to visit a shrine, then to walk 300 meters east to visit a temple.

Query 2: If we start from a point at a distance of 2000 meters from the west end of the road, the optimal move is first to walk 1000 meters west to visit a temple, then to walk 400 meters west to visit a shrine. We will pass by another temple on the way, but it is fine.

Query 3: If we start from a point at a distance of 899 meters from the west end of the road, the optimal move is first to walk 1 meter east to visit a temple, then to walk 300 meters west to visit a shrine.

Query 4: If we start from a point at a distance of 799 meters from the west end of the road, the optimal move is first to walk 199 meters west to visit a shrine, then to walk 200 meters west to visit a temple.

Sample Input 2

1 1 3
1
10000000000
2
9999999999
5000000000

Sample Output 2

10000000000
10000000000
14999999998

The road is quite long, and we may need to travel a distance that does not fit into a 32-bit integer.



Sample input:
2 3 4
100
600
400
900
1000
150
2000
899
799


Sample output:
350
1400
301
399

Submission status: Accepted

Solution code:
import bisect
A, B, Q = map(int, input().split())
S = [int(input()) for _ in range(A)]
T = [int(input()) for _ in range(B)]
X = [int(input()) for _ in range(Q)]

for x in X:
    p0, p1, p2, p3 = (10**12 for _ in range(4))
    si = bisect.bisect_left(S, x)
    ti = bisect.bisect_left(T, x)

    if si != 0 and ti != 0:
        p0 = max(x-S[si-1], x-T[ti-1])
    if si != 0 and ti != B:
        p1 = min(x-S[si-1], T[ti]-x)*2+max(x-S[si-1], T[ti]-x)
    if si != A and ti != 0:
        p2 = min(S[si]-x, x-T[ti-1])*2+max(S[si]-x, x-T[ti-1])
    if si != A and ti != B:
        p3 = max(S[si]-x, T[ti]-x)
    print(min(p0, p1, p2, p3))

Submission status: Accepted

Solution code:
import bisect
A,B,Q=map(int,input().split())
inf=10**18
s=[-inf]+[int(input()) for i in range(A)]+[inf]
t=[-inf]+[int(input()) for i in range(B)]+[inf]

for i in range(Q):
  ans=inf
  x=int(input())
  b=bisect.bisect_left(s,x)
  d=bisect.bisect_left(t,x)
  for S in (s[b-1],s[b]):
    for T in (t[d-1],t[d]):
      ans=min(ans,abs(S-x)+abs(S-T),abs(T-x)+abs(S-T))
  print(ans)

Submission status: Accepted

Solution code:
import math
import bisect
A,B,Q = list(map(int,input().split(" ")))
Ss = [int(input())for _ in range(A)]
Ts = [int(input())for _ in range(B)]
Xs = [int(input())for _ in range(Q)]
Ss.sort()
Ts.sort()
ss = len(Ss)
ts = len(Ts)
for x in Xs:
    si = bisect.bisect_right(Ss,x)
    ti = bisect.bisect_right(Ts,x)

    cands = [[],[]]
    if si > 0:
        cands[0].append(Ss[si-1])
    if si < ss:
        cands[0].append(Ss[si])
    if ti > 0:
        cands[1].append(Ts[ti-1])
    if ti < ts:
        cands[1].append(Ts[ti])
 
    anss = []
    temp = 0
    for i in [0,1]:
        for t1 in cands[i]: #最初による所
            for t2 in cands[1-i]: #次に寄る所
                temp = abs(x-t1) +  abs(t2-t1)
                anss.append(temp)
    print(min(anss))

Submission status: Accepted

Solution code:
from bisect import bisect_left as bl
A, B, Q = map(int, input().split())
inf = 10 ** 11
Sh = [-1 * inf] + [int(input()) for _ in range(A)] + [inf]
Te = [-1 * inf] + [int(input()) for _ in range(B)] + [inf]
for _ in range(Q):
  x = int(input())
  b, d = bl(Sh, x), bl(Te, x)
  res = inf
  for S in [Sh[b - 1], Sh[b]]:
    for T in [Te[d - 1], Te[d]]:
      d1, d2 = abs(S - x) + abs(T - S), abs(T - x) + abs(S - T)
      res = min(res, d1, d2)
  print(res)

Submission status: Accepted

Solution code:
# ABC 119 D

import bisect

def resolve():
    A, B, Q = map(int, input().split())
    s = [int(input()) for _ in range(A)]
    t = [int(input()) for _ in range(B)]

    for _ in range(Q):
        i = int(input())
        s1 = -1 if bisect.bisect_left(s, i) == 0 else s[bisect.bisect_left(s, i)-1]
        s2 = -1 if bisect.bisect_right(s, i) >= A else s[bisect.bisect_right(s, i)]
        t1 = -1 if bisect.bisect_left(t, i) == 0 else t[bisect.bisect_left(t, i)-1]
        t2 = -1 if bisect.bisect_left(t, i) >= B else t[bisect.bisect_right(t, i)]
        tmp = []

        if s1 != -1 and t1 != -1:
            tmp.append(max(i-s1, i-t1))
        if s2 != -1 and t2 != -1:
            tmp.append(max(s2-i, t2-i))
        if s1 != -1 and t2 != -1:
            tmp.append(min(i-s1, t2-i) + t2-s1)
        if t1 != -1 and s2 != -1:
            tmp.append(min(i-t1, s2-i) + s2-t1)

        print(min(tmp))


if __name__ == "__main__":
    resolve()

Submission status: Accepted

Solution code:
from bisect import bisect_left, bisect_right

INF = int(1E18)
A, B, Q = map(int, input().split())
S = [int(input()) for _ in range(A)]
T = [int(input()) for _ in range(B)]
X = [int(input()) for _ in range(Q)]
S.insert(0, -INF)
T.insert(0, -INF)
S.append(INF)
T.append(INF)

for x in X:
    i = bisect_left(S, x)
    j = bisect_left(T, x)
    s1, s2 = S[i-1], S[i]
    t1, t2 = T[j-1], T[j]
    s1 = abs(s1-x)
    s2 = abs(s2-x)
    t1 = abs(t1-x)
    t2 = abs(t2-x)
    l1 = max(s1, t1)
    l2 = min(s1, t1)
    r1 = max(s2, t2)
    r2 = min(s2, t2)
    E = [l1, r1]
    if (s1-t1)*(s2-t2) < 0:
        E.extend([2*l2+r2, 2*r2+l2])
    print(min(E))

Submission status: Accepted

Solution code:
def main():

    """ convenient functions
    # for i, a in enumerate(iterable)
    # q, mod = divmod(a, b)
    # divmod(x, y) returns the tuple (x//y, x%y)
    # Higher-order function: reduce(operator.mul, xyz_count, 1)
    # manage median(s) using two heapq https://atcoder.jp/contests/abc127/tasks/abc127_f
    """

    """convenient decorator
    # @functools.lru_cache():
    # to facilitate use of recursive function
        # ex:
        # from functools import lru_cache
        # import sys
        # sys.setrecursionlimit(10**9)
        # @lru_cache(maxsize=None)
        # def fib(n):
        #     if n < 2:
        #         return n
        #     return fib(n-1) + fib(n-2)
        # print(fib(1000))
    """
    
    # import numpy as np
    import sys
    sys.setrecursionlimit(10**7)
    from itertools import accumulate, combinations, permutations, product # https://docs.python.org/ja/3/library/itertools.html
    # accumulate() returns iterator! to get list: list(accumulate())
    from math import factorial, ceil, floor, sqrt
    def factorize(n):
        """return the factors of the Arg and count of each factor
        
        Args:
            n (long): number to be resolved into factors
        
        Returns:
            list of tuples: factorize(220) returns [(2, 2), (5, 1), (11, 1)]
        """
        fct = []  # prime factor
        b, e = 2, 0  # base, exponent
        while b * b <= n:
            while n % b == 0:
                n = n // b
                e = e + 1
            if e > 0:
                fct.append((b, e))
            b, e = b + 1, 0
        if n > 1:
            fct.append((n, 1))
        return fct
    def combinations_count(n, r):
        """Return the number of selecting r pieces of items from n kinds of items.
        
        Args:
            n (long): number
            r (long): number
        
        Raises:
            Exception: not defined when n or r is negative
        
        Returns:
            long: number
        """
        # TODO: How should I do when n - r is negative?
        if n < 0 or r < 0:
            raise Exception('combinations_count(n, r) not defined when n or r is negative')
        if n - r < r: r = n - r
        if r < 0: return 0
        if r == 0: return 1
        if r == 1: return n
        numerator = [n - r + k + 1 for k in range(r)]
        denominator = [k + 1 for k in range(r)]
        for p in range(2,r+1):
            pivot = denominator[p - 1]
            if pivot > 1:
                offset = (n - r) % p
                for k in range(p-1,r,p):
                    numerator[k - offset] /= pivot
                    denominator[k] /= pivot
        result = 1
        for k in range(r):
            if numerator[k] > 1:
                result *= int(numerator[k])
        return result
    def combinations_with_replacement_count(n, r):
        """Return the number of selecting r pieces of items from n kinds of items allowing individual elements to be repeated more than once.
        
        Args:
            n (long): number
            r (long): number
        
        Raises:
            Exception: not defined when n or r is negative
        
        Returns:
            long: number
        """
        if n < 0 or r < 0:
            raise Exception('combinations_with_replacement_count(n, r) not defined when n or r is negative')
        elif n == 0:
            return 1
        else:
            return combinations_count(n + r - 1, r)
    from bisect import bisect_left, bisect_right
    from collections import deque, Counter, defaultdict # https://docs.python.org/ja/3/library/collections.html#collections.deque
    from heapq import heapify, heappop, heappush, heappushpop, heapreplace,nlargest,nsmallest # https://docs.python.org/ja/3/library/heapq.html
    from copy import deepcopy, copy # https://docs.python.org/ja/3/library/copy.html
    import operator
    from operator import itemgetter
    # ex1: List.sort(key=itemgetter(1))
    # ex2: sorted(tuples, key=itemgetter(1,2))
    from functools import reduce, lru_cache
    def chmin(x, y):
        """change minimum
        if x > y, x = y and return (x, True).
        convenient when solving problems of dp[i]
        
        Args:
            x (long): current minimum value
            y (long): potential minimum value
        
        Returns:
            (x, bool): (x, True) when updated, else (x, False)
        """
        if x > y:
            x = y
            return (x, True)
        else:
            return (x, False)
    def chmax(x, y):
        """change maximum
        if x < y, x = y and return (x, True).
        convenient when solving problems of dp[i]
        
        Args:
            x (long): current maximum value
            y (long): potential maximum value
        
        Returns:
            (x, bool): (x, True) when updated, else (x, False)
        """
        if x < y:
            x = y
            return (x, True)
        else:
            return (x, False)

    from fractions import gcd # Deprecated since version 3.5: Use math.gcd() instead.
    def gcds(numbers):
        return reduce(gcd, numbers)
    def lcm(x, y):
        return (x * y) // gcd(x, y)
    def lcms(numbers):
        return reduce(lcm, numbers, 1)

    # first create factorial_list
    # fac_list = mod_factorial_list(n)
    INF = 10 ** 18
    MOD = 10 ** 9 + 7
    modpow = lambda a, n, p = MOD: pow(a, n, p) # Recursive function in python is slow!
    def modinv(a, p = MOD):
        # evaluate reciprocal using Fermat's little theorem:
        # a**(p-1) is identical to 1 (mod p) when a and p is coprime
        return modpow(a, p-2, p)
    def modinv_list(n, p = MOD):
        if n <= 1:
            return [0,1][:n+1]
        else:
            inv_t = [0,1]
            for i in range(2, n+1):
                inv_t += [inv_t[p % i] * (p - int(p / i)) % p]
            return inv_t
    def modfactorial_list(n, p = MOD):
        if n == 0:
            return [1]
        else:
            l = [0] * (n+1)
            tmp = 1
            for i in range(1, n+1):
                tmp = tmp * i % p
                l[i] = tmp
            return l
    def modcomb(n, k, fac_list = [], p = MOD):
        # fac_list = modfactorial_list(100)
        # print(modcomb(100, 5, modfactorial_list(100)))
        from math import factorial
        if n < 0 or k < 0 or n < k: return 0
        if n == 0 or k == 0: return 1
        if len(fac_list) <= n:
            a = factorial(n) % p
            b = factorial(k) % p
            c = factorial(n-k) % p
        else:
            a = fac_list[n]
            b = fac_list[k]
            c = fac_list[n-k]
        return (a * modpow(b, p-2, p) * modpow(c, p-2, p)) % p
    def modadd(a, b, p = MOD):
        return (a + b) % MOD
    def modsub(a, b, p = MOD):
        return (a - b) % p
    def modmul(a, b, p = MOD):
        return ((a % p) * (b % p)) % p
    def moddiv(a, b, p = MOD):
        return modmul(a, modpow(b, p-2, p))

    class UnionFindTree:
        """union find tree class
        TODO: fix this description...
        how to use (example):
        >>  uf = UnionFindTree(N) 
        >>  if uf.find_root(a) == uf.find_root(b):
        >>      do something
        >>  else:
        >>      do something
        >>      uf.unite(a, b)
        """
        def __init__(self, N):
            self.root = [-1] * (N+1)
            self.rank   = [0]  * (N+1)
            self.connected_num = [1] * (N+1)
    
        def find_root(self,x):
            root = self.root
            while root[x] != -1:
                x = root[x]
            return x
    
        def unite(self,x,y):
            root = self.root
            rank = self.rank
            connected_num = self.connected_num
            find_root = self.find_root
    
            rx = find_root(x)
            ry = find_root(y)
            if rx != ry:
                if rank[rx] < rank[ry]:
                    root[rx] = ry
                    rx,ry = ry,rx
                else:
                    if rank[rx] == rank[ry]:
                        rank[rx] += 1
                    root[ry] = rx       
                connected_num[rx] += connected_num[ry]

    """ initialize variables and set inputs
    # initialize variables
        # to initialize list, use [0] * n
        # to initialize two dimentional array, use [[0] * N for _ in range(N)]
    # set inputs
        # open(0).read() is a convenient method:
        # ex) n, m, *x = map(int, open(0).read().split())
        #     min(x[::2]) - max(x[1::2])
        # ex2) *x, = map(int, open(0).read().split())
        #     don't forget to add comma after *x if only one variable is used
    # preprocessing
        # transpose = [x for x in zip(*data)]
        # ex) [[1, 2, 3], [4, 5, 6], [7, 8, 9]] => [(1, 4, 7), (2, 5, 8), (3, 6, 9)]
        # flat = [flatten for inner in data for flatten in inner]
        # ex) [[1, 2, 3], [4, 5, 6], [7, 8, 9]] => [1, 2, 3, 4, 5, 6, 7, 8, 9]
    # calculate and output
        # output pattern
        # ex1) print(*l) => when l = [2, 5, 6], printed 2 5 6
    """

    # functions used
    r = lambda: sys.stdin.readline().strip()
    r_int = lambda: int(r())
    R = lambda: list(map(int, r().split()))
    Rmap = lambda: map(int, r().split())
    Rfloat = lambda: list(map(float, r().split()))
    Rtuple = lambda: tuple(map(int, r().split()))

    """ how to treat input
    # single int: int(r())
    # single string: r()
    # single float: float(r())
    # line int: R()
    # line string: r().split()
    # line (str, int, int): [j if i == 0 else int(j) for i, j in enumerate(r().split())]
    # lines int: [R() for _ in range(n)]
    """

    if __name__ != '__main__':
        # for test
        sys.stdin = open('sample.txt')
        
    # ----------------------------------
    # main

    A, B, Q = R()
    S = [-INF] + [int(r()) for _ in range(A)] + [INF] 
    T = [-INF] + [int(r()) for _ in range(B)] + [INF] 
    X = [int(r()) for _ in range(Q)]
    for x in X:
        s = bisect_left(S, x)
        t = bisect_left(T, x)
        # if s == 0:
        #     sl = -INF
        #     sr = S[s]
        # elif s == A:
        #     sl = S[s-1]
        #     sr = INF
        # else:
        sl = S[s-1]
        sr = S[s]
        # if t == 0:
        #     tl = -INF
        #     tr = T[t]
        # elif t == B:
        #     tl = T[t-1]
        #     tr = INF
        # else:
        tl = T[t-1]
        tr = T[t]
        print(min(min(abs(x-sl), abs(x-tr))+abs(sl-tr),
            min(abs(x-sr), abs(x-tl))+abs(sr-tl),
            max(abs(x-sl), abs(x-tl)),
            max(abs(x-sr), abs(x-tr)))
              )
        
    
    # end of main
    # ----------------------------------

    """memo: how to use defaultdict of list
    # initialize
    Dic = defaultdict(list)
    # append / extend
    Dic[x].append(y)
    # three methods for loop: keys(), values(), items()
    for k, v in Dic.items():
    """
    
    """memo: how to solve binary problems
    # to make binary digits text
    >>> a = 5
    >>> bin_str_a = format(a, '#06b')
    >>> print(bin_str_a)
    0b0101 # first 2 strings (='0b') indicates it is binary
    """
    
    """memo: how to solve the problem
    creating simple test/answer
    greed
    simple dp
    graph
    """

if __name__ == '__main__':
    main()

Submission status: Accepted

Solution code:
def main():

    """ convenient functions
    # for i, a in enumerate(iterable)
    # q, mod = divmod(a, b)
    # divmod(x, y) returns the tuple (x//y, x%y)
    # Higher-order function: reduce(operator.mul, xyz_count, 1)
    # manage median(s) using two heapq https://atcoder.jp/contests/abc127/tasks/abc127_f
    """

    """convenient decorator
    # @functools.lru_cache():
    # to facilitate use of recursive function
        # ex:
        # from functools import lru_cache
        # import sys
        # sys.setrecursionlimit(10**9)
        # @lru_cache(maxsize=None)
        # def fib(n):
        #     if n < 2:
        #         return n
        #     return fib(n-1) + fib(n-2)
        # print(fib(1000))
    """
    
    # import numpy as np
    import sys
    sys.setrecursionlimit(10**7)
    from itertools import accumulate, combinations, permutations, product # https://docs.python.org/ja/3/library/itertools.html
    # accumulate() returns iterator! to get list: list(accumulate())
    from math import factorial, ceil, floor, sqrt
    def factorize(n):
        """return the factors of the Arg and count of each factor
        
        Args:
            n (long): number to be resolved into factors
        
        Returns:
            list of tuples: factorize(220) returns [(2, 2), (5, 1), (11, 1)]
        """
        fct = []  # prime factor
        b, e = 2, 0  # base, exponent
        while b * b <= n:
            while n % b == 0:
                n = n // b
                e = e + 1
            if e > 0:
                fct.append((b, e))
            b, e = b + 1, 0
        if n > 1:
            fct.append((n, 1))
        return fct
    def combinations_count(n, r):
        """Return the number of selecting r pieces of items from n kinds of items.
        
        Args:
            n (long): number
            r (long): number
        
        Raises:
            Exception: not defined when n or r is negative
        
        Returns:
            long: number
        """
        # TODO: How should I do when n - r is negative?
        if n < 0 or r < 0:
            raise Exception('combinations_count(n, r) not defined when n or r is negative')
        if n - r < r: r = n - r
        if r < 0: return 0
        if r == 0: return 1
        if r == 1: return n
        numerator = [n - r + k + 1 for k in range(r)]
        denominator = [k + 1 for k in range(r)]
        for p in range(2,r+1):
            pivot = denominator[p - 1]
            if pivot > 1:
                offset = (n - r) % p
                for k in range(p-1,r,p):
                    numerator[k - offset] /= pivot
                    denominator[k] /= pivot
        result = 1
        for k in range(r):
            if numerator[k] > 1:
                result *= int(numerator[k])
        return result
    def combinations_with_replacement_count(n, r):
        """Return the number of selecting r pieces of items from n kinds of items allowing individual elements to be repeated more than once.
        
        Args:
            n (long): number
            r (long): number
        
        Raises:
            Exception: not defined when n or r is negative
        
        Returns:
            long: number
        """
        if n < 0 or r < 0:
            raise Exception('combinations_with_replacement_count(n, r) not defined when n or r is negative')
        elif n == 0:
            return 1
        else:
            return combinations_count(n + r - 1, r)
    from bisect import bisect_left, bisect_right
    from collections import deque, Counter, defaultdict # https://docs.python.org/ja/3/library/collections.html#collections.deque
    from heapq import heapify, heappop, heappush, heappushpop, heapreplace,nlargest,nsmallest # https://docs.python.org/ja/3/library/heapq.html
    from copy import deepcopy, copy # https://docs.python.org/ja/3/library/copy.html
    import operator
    from operator import itemgetter
    # ex1: List.sort(key=itemgetter(1))
    # ex2: sorted(tuples, key=itemgetter(1,2))
    from functools import reduce, lru_cache
    def chmin(x, y):
        """change minimum
        if x > y, x = y and return (x, True).
        convenient when solving problems of dp[i]
        
        Args:
            x (long): current minimum value
            y (long): potential minimum value
        
        Returns:
            (x, bool): (x, True) when updated, else (x, False)
        """
        if x > y:
            x = y
            return (x, True)
        else:
            return (x, False)
    def chmax(x, y):
        """change maximum
        if x < y, x = y and return (x, True).
        convenient when solving problems of dp[i]
        
        Args:
            x (long): current maximum value
            y (long): potential maximum value
        
        Returns:
            (x, bool): (x, True) when updated, else (x, False)
        """
        if x < y:
            x = y
            return (x, True)
        else:
            return (x, False)

    from fractions import gcd # Deprecated since version 3.5: Use math.gcd() instead.
    def gcds(numbers):
        return reduce(gcd, numbers)
    def lcm(x, y):
        return (x * y) // gcd(x, y)
    def lcms(numbers):
        return reduce(lcm, numbers, 1)

    # first create factorial_list
    # fac_list = mod_factorial_list(n)
    INF = 10 ** 18
    MOD = 10 ** 9 + 7
    modpow = lambda a, n, p = MOD: pow(a, n, p) # Recursive function in python is slow!
    def modinv(a, p = MOD):
        # evaluate reciprocal using Fermat's little theorem:
        # a**(p-1) is identical to 1 (mod p) when a and p is coprime
        return modpow(a, p-2, p)
    def modinv_list(n, p = MOD):
        if n <= 1:
            return [0,1][:n+1]
        else:
            inv_t = [0,1]
            for i in range(2, n+1):
                inv_t += [inv_t[p % i] * (p - int(p / i)) % p]
            return inv_t
    def modfactorial_list(n, p = MOD):
        if n == 0:
            return [1]
        else:
            l = [0] * (n+1)
            tmp = 1
            for i in range(1, n+1):
                tmp = tmp * i % p
                l[i] = tmp
            return l
    def modcomb(n, k, fac_list = [], p = MOD):
        # fac_list = modfactorial_list(100)
        # print(modcomb(100, 5, modfactorial_list(100)))
        from math import factorial
        if n < 0 or k < 0 or n < k: return 0
        if n == 0 or k == 0: return 1
        if len(fac_list) <= n:
            a = factorial(n) % p
            b = factorial(k) % p
            c = factorial(n-k) % p
        else:
            a = fac_list[n]
            b = fac_list[k]
            c = fac_list[n-k]
        return (a * modpow(b, p-2, p) * modpow(c, p-2, p)) % p
    def modadd(a, b, p = MOD):
        return (a + b) % MOD
    def modsub(a, b, p = MOD):
        return (a - b) % p
    def modmul(a, b, p = MOD):
        return ((a % p) * (b % p)) % p
    def moddiv(a, b, p = MOD):
        return modmul(a, modpow(b, p-2, p))

    class UnionFindTree:
        """union find tree class
        TODO: fix this description...
        how to use (example):
        >>  uf = UnionFindTree(N) 
        >>  if uf.find_root(a) == uf.find_root(b):
        >>      do something
        >>  else:
        >>      do something
        >>      uf.unite(a, b)
        """
        def __init__(self, N):
            self.root = [-1] * (N+1)
            self.rank   = [0]  * (N+1)
            self.connected_num = [1] * (N+1)
    
        def find_root(self,x):
            root = self.root
            while root[x] != -1:
                x = root[x]
            return x
    
        def unite(self,x,y):
            root = self.root
            rank = self.rank
            connected_num = self.connected_num
            find_root = self.find_root
    
            rx = find_root(x)
            ry = find_root(y)
            if rx != ry:
                if rank[rx] < rank[ry]:
                    root[rx] = ry
                    rx,ry = ry,rx
                else:
                    if rank[rx] == rank[ry]:
                        rank[rx] += 1
                    root[ry] = rx       
                connected_num[rx] += connected_num[ry]

    """ initialize variables and set inputs
    # initialize variables
        # to initialize list, use [0] * n
        # to initialize two dimentional array, use [[0] * N for _ in range(N)]
    # set inputs
        # open(0).read() is a convenient method:
        # ex) n, m, *x = map(int, open(0).read().split())
        #     min(x[::2]) - max(x[1::2])
        # ex2) *x, = map(int, open(0).read().split())
        #     don't forget to add comma after *x if only one variable is used
    # preprocessing
        # transpose = [x for x in zip(*data)]
        # ex) [[1, 2, 3], [4, 5, 6], [7, 8, 9]] => [(1, 4, 7), (2, 5, 8), (3, 6, 9)]
        # flat = [flatten for inner in data for flatten in inner]
        # ex) [[1, 2, 3], [4, 5, 6], [7, 8, 9]] => [1, 2, 3, 4, 5, 6, 7, 8, 9]
    # calculate and output
        # output pattern
        # ex1) print(*l) => when l = [2, 5, 6], printed 2 5 6
    """

    # functions used
    r = lambda: sys.stdin.readline().strip()
    r_int = lambda: int(r())
    R = lambda: list(map(int, r().split()))
    Rmap = lambda: map(int, r().split())
    Rfloat = lambda: list(map(float, r().split()))
    Rtuple = lambda: tuple(map(int, r().split()))

    """ how to treat input
    # single int: int(r())
    # single string: r()
    # single float: float(r())
    # line int: R()
    # line string: r().split()
    # line (str, int, int): [j if i == 0 else int(j) for i, j in enumerate(r().split())]
    # lines int: [R() for _ in range(n)]
    """

    if __name__ != '__main__':
        # for test
        sys.stdin = open('sample.txt')
        
    # ----------------------------------
    # main

    A, B, Q = R()
    S = [int(r()) for _ in range(A)]
    T = [int(r()) for _ in range(B)]
    X = [int(r()) for _ in range(Q)]
    for x in X:
        s = bisect_left(S, x)
        t = bisect_left(T, x)
        if s == 0:
            sl = -INF
            sr = S[s]
        elif s == A:
            sl = S[s-1]
            sr = INF
        else:
            sl = S[s-1]
            sr = S[s]
        if t == 0:
            tl = -INF
            tr = T[t]
        elif t == B:
            tl = T[t-1]
            tr = INF
        else:
            tl = T[t-1]
            tr = T[t]
        print(min(min(abs(x-sl), abs(x-tr))+abs(sl-tr),
            min(abs(x-sr), abs(x-tl))+abs(sr-tl),
            max(abs(x-sl), abs(x-tl)),
            max(abs(x-sr), abs(x-tr)))
              )
        
    
    # end of main
    # ----------------------------------

    """memo: how to use defaultdict of list
    # initialize
    Dic = defaultdict(list)
    # append / extend
    Dic[x].append(y)
    # three methods for loop: keys(), values(), items()
    for k, v in Dic.items():
    """
    
    """memo: how to solve binary problems
    # to make binary digits text
    >>> a = 5
    >>> bin_str_a = format(a, '#06b')
    >>> print(bin_str_a)
    0b0101 # first 2 strings (='0b') indicates it is binary
    """
    
    """memo: how to solve the problem
    creating simple test/answer
    greed
    simple dp
    graph
    """

if __name__ == '__main__':
    main()

Submission status: Accepted

Solution code:
from bisect import bisect as B
I=lambda z:[int(input()) for _ in range(z)]
a,b,q=map(int,input().split())
F=float("inf")
s=[-F]+I(a)+[F];t=[-F]+I(b)+[F];X=I(q)
for x in X:
	i=B(s,x);j=B(t,x)
	o,p=x-s[i-1],s[i]-x
	q,r=x-t[j-1],t[j]-x
	print(min(2*o+r,o+r*2,2*p+q,p+q*2,max(o,q),max(p,r)))

Submission status: Accepted

Solution code:
from bisect import bisect
a,b,q=map(int,input().split())
INF=float("inf")
s=[-INF]+[int(input()) for _ in range(a)]+[INF]
t=[-INF]+[int(input()) for _ in range(b)]+[INF]
xx=[int(input()) for _ in range(q)]
for x in xx:
	si=bisect(s,x)
	ti=bisect(t,x)
	sf,sb=x-s[si-1],s[si]-x
	tf,tb=x-t[ti-1],t[ti]-x
	print(min(2*sf+tb,sf+tb*2,2*sb+tf,sb+tf*2,max(sf,tf),max(sb,tb)))

Submission status: Accepted

Solution code:
import bisect
A, B, Q = map(int, input().split())
INF = 10 ** 18
s = [-INF] + [int(input()) for i in range(A)] + [INF]
t = [-INF] + [int(input()) for i in range(B)] + [INF]
for q in range(Q):
  x = int(input())
  b, d = bisect.bisect_right(s, x), bisect.bisect_right(t, x)
  res = INF
  for S in [s[b - 1], s[b]]:
    for T in [t[d - 1], t[d]]:
      d1, d2 = abs(S - x) + abs(T - S), abs(T - x) + abs(S - T)
      res = min(res, d1, d2)
  print(res)

Submission status: Accepted

Solution code:
from bisect import bisect_right

A, B, Q = map(int, input().split())
s = [int(input()) for _ in range(A)]
t = [int(input()) for _ in range(B)]
x = [int(input()) for _ in range(Q)]

INF = 10 ** 18
s = [-INF] + s + [INF]
t = [-INF] + t + [INF]

for i in range(Q):
    b = bisect_right(s, x[i])
    d = bisect_right(t, x[i])
    res = INF
    for n in [s[b - 1], s[b]]:
        for m in [t[d - 1], t[d]]:
            d1 = abs(n - x[i]) + abs(m - n)
            d2 = abs(m - x[i]) + abs(n - m)
            res = min(res, d1, d2)
    print(res)

Submission status: Accepted

Solution code:
import bisect
INF = 10 ** 18
a, b, Q = map(int, input().split())
s = [-INF] + [int(input()) for i in range(a)] + [INF]
t = [-INF] + [int(input()) for i in range(b)] + [INF]
for q in range(Q):
  x = int(input())
  s_idx = bisect.bisect_left(s, x)
  t_idx = bisect.bisect_left(t, x)
  ans = INF
  for i in [s_idx, s_idx-1]:
    for j in [t_idx, t_idx-1]:
      temp1 = abs(s[i] - x) + abs(s[i] - t[j])
      temp2 = abs(t[j] - x) + abs(s[i] - t[j])
      ans = min(ans, temp1, temp2)
  print(ans)