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ABC328をPythonで(A~F)

Posted at

トヨタ自動車プログラミングコンテスト2023#7(AtCoder Beginner Contest 328)

A問題

問題文の通りにやる。

A
n, x = map(int, input().split())
a = list(map(int, input().split()))
print(sum(a_i for a_i in a if a_i <= x))

B問題

全部調べる。

B
n = int(input())
d = list(map(int, input().split()))
ans = 0
for i, d_i in enumerate(d, 1):
    for j in range(1, d_i + 1):
        if len(set(list(str(i) + str(j)))) == 1:
            ans += 1

print(ans)

C問題

i番目までに連続しているところがいくつあるか事前に数える。

C
n, q = map(int, input().split())
s = input()

d = [0] * n
for i in range(1, n):
    d[i] += d[i - 1]
    if s[i] == s[i - 1]:
        d[i] += 1

for _ in range(q):
    l, r = map(int, input().split())
    print(d[r - 1] - d[l - 1])

D問題

先頭から貪欲にやっていく。

D
s = input()

ans = []
for s_i in s:
    ans.append(s_i)
    if len(ans) >= 3 and ans[-3:] == ["A", "B", "C"]:
        ans.pop();ans.pop();ans.pop()

print("".join(ans))

E問題

辺を$n-1$本選んですべて連結するか確認。
その中で最小のものを出力。

E
from itertools import combinations
from collections import defaultdict

class UnionFind:
    """
    URL : https://note.nkmk.me/python-union-find/
    """
    def __init__(self, n):
        self.n = n
        self.parents = [-1] * n

    def find(self, x):
        if self.parents[x] < 0:
            return x
        else:
            self.parents[x] = self.find(self.parents[x])
            return self.parents[x]

    def union(self, x, y):
        x = self.find(x)
        y = self.find(y)

        if x == y:
            return

        if self.parents[x] > self.parents[y]:
            x, y = y, x

        self.parents[x] += self.parents[y]
        self.parents[y] = x

    def size(self, x):
        return -self.parents[self.find(x)]

    def same(self, x, y):
        return self.find(x) == self.find(y)

    def members(self, x):
        root = self.find(x)
        return [i for i in range(self.n) if self.find(i) == root]

    def roots(self):
        return [i for i, x in enumerate(self.parents) if x < 0]

    def group_count(self):
        return len(self.roots())


n, m, k = map(int, input().split())
data = []
for _ in range(m):
    u, v, w = map(int, input().split())
    data.append((u - 1, v - 1, w))

ans = float("inf")
for c in combinations(range(m), n - 1):
    UF = UnionFind(n)
    ans_i = 0
    for c_i in c:
        u, v, w = data[c_i]
        UF.union(u, v)
        ans_i = (ans_i + w) % k
    
    if UF.group_count() == 1:
        ans = min(ans, ans_i)

print(ans)

F問題

重み付きUnionFindで順にやっていくだけ。

F
class WeightedUnionFind:
    """
    URL:https://at274.hatenablog.com/entry/2018/02/03/140504#%E5%AE%8C%E6%88%90%E5%BD%A2
    """
    def __init__(self, n):
        self.par = [i for i in range(n+1)]
        self.rank = [0] * (n+1)
        # 根への距離を管理
        self.weight = [0] * (n+1)

    # 検索
    def find(self, x):
        if self.par[x] == x:
            return x
        else:
            y = self.find(self.par[x])
            # 親への重みを追加しながら根まで走査
            self.weight[x] += self.weight[self.par[x]]
            self.par[x] = y
            return y

    # 併合
    def union(self, x, y, w):
        rx = self.find(x)
        ry = self.find(y)
        # xの木の高さ < yの木の高さ
        if self.rank[rx] < self.rank[ry]:
            self.par[rx] = ry
            self.weight[rx] = w - self.weight[x] + self.weight[y]
        # xの木の高さ ≧ yの木の高さ
        else:
            self.par[ry] = rx
            self.weight[ry] = -w - self.weight[y] + self.weight[x]
            # 木の高さが同じだった場合の処理
            if self.rank[rx] == self.rank[ry]:
                self.rank[rx] += 1

    # 同じ集合に属するか
    def same(self, x, y):
        return self.find(x) == self.find(y)

    # xからyへのコスト
    def diff(self, x, y):
        return self.weight[x] - self.weight[y]


n, q = map(int, input().split())
ans = []
UF = WeightedUnionFind(n + 10)
for i in range(1, q + 1):
    a, b, d = map(int, input().split())
    if UF.same(a, b) and UF.diff(a, b) != d:
        continue
    else:
        ans.append(i)
        UF.union(a, b, d)

print(*ans)
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