abc338

Author: reo11

atcoder/abc/abc301-400/abc338/a.py

@@ -0,0 +1,17 @@
+s = list(input())
+
+flag = True
+big_alphabet = set([chr(ord("A") + i) for i in range(26)])
+small_alphabet = set([chr(ord("a") + i) for i in range(26)])
+
+if s[0] not in big_alphabet:
+    flag = False
+
+for i in range(1, len(s)):
+    if s[i] not in small_alphabet:
+        flag = False
+
+if flag:
+    print("Yes")
+else:
+    print("No")

atcoder/abc/abc301-400/abc338/b.py

@@ -0,0 +1,16 @@
+from collections import defaultdict
+
+counter = defaultdict(int)
+
+s = list(input())
+
+for si in s:
+    counter[si] += 1
+
+ans = [-1, ""]
+for i in range(26):
+    si = chr(ord("a") + i)
+    cnt = counter[si]
+    if ans[0] < cnt:
+        ans = [cnt, si]
+print(ans[1])

atcoder/abc/abc301-400/abc338/c.py

@@ -0,0 +1,23 @@
+n = int(input())
+q = list(map(int, input().split()))
+a = list(map(int, input().split()))
+b = list(map(int, input().split()))
+
+def solve(q, x):
+    # 材料qでxを作れる最大数
+    ans = []
+    for qi, xi in zip(q, x):
+        if xi == 0:
+            continue
+        ans.append(qi // xi)
+    return min(ans)
+
+# 全探索
+ans = 0
+max_a = solve(q, a)
+for i in range(0, max_a + 1):
+    q_tmp = [qi - i * xi for qi, xi in zip(q, a)]
+    max_b = solve(q_tmp, b)
+    ans = max(ans, i + max_b)
+print(ans)
+

atcoder/abc/abc301-400/abc338/d.py

@@ -0,0 +1,41 @@
+from collections import deque, Counter
+n, m = map(int, input().split())
+x = list(map(int, input().split()))
+
+# xと(x+1)mod nを繋いでいるので経路は時計回りか反時計回り
+# 最初に適当に最短になるように移動させて、最も通行回数が少ない辺を削除する
+# 通行回数が少ない辺が複数ある場合は全部試す？（未証明だが一旦1箇所で試す）
+
+# 累積和の差分を格納
+counter = [0] * (2 * (n + 1))
+q = deque(x)
+
+current_num = q.popleft()
+cost = 0
+while q:
+    next_num = q.popleft()
+    max_num = max(current_num, next_num)
+    min_num = min(current_num, next_num)
+    # 時計回り
+    diff1 = max_num - min_num
+    # 反時計回り
+    diff2 = (n + min_num) - max_num
+    if diff1 < diff2:
+        counter[min_num] += diff2 - diff1
+        counter[max_num] -= diff2 - diff1
+    elif diff1 > diff2:
+        counter[max_num] += diff1 - diff2
+        counter[min_num + n] -= diff1 - diff2
+
+    cost += min(diff1, diff2)
+    # print(cost, current_num, next_num, diff1, diff2)
+    current_num = next_num
+# print(counter)
+# 累積和
+losses = [0] * (2 * (n + 1))
+for i in range(1, 2 * (n + 1)):
+    losses[i] += losses[i - 1] + counter[i]
+
+for i in range(1, n + 1):
+    losses[i] = losses[i] + losses[n + i]
+print(cost + min(losses[1:n + 1]))

atcoder/abc/abc301-400/abc338/e.py

@@ -0,0 +1,25 @@
+from collections import deque
+n = int(input())
+ab = []
+stuck = deque()
+
+for i in range(n):
+    a, b = map(int, input().split())
+    min_ab = min(a, b)
+    max_ab = max(a, b)
+    ab.append([min_ab, "a", i])
+    ab.append([max_ab, "b", i])
+ab.sort()
+
+ans = False
+for v, is_ab, i in ab:
+    if is_ab == "a":
+        stuck.append(i)
+    else:
+        if len(stuck) > 0 and stuck[-1] == i:
+            stuck.pop()
+        else:
+            ans = True
+            break
+
+print("Yes" if ans else "No")

atcoder/abc/abc301-400/abc338/f.py

@@ -0,0 +1,44 @@
+INF = float('inf')
+n, m = map(int, input().split())
+G = [[INF] * n for _ in range(n)]
+
+for _ in range(m):
+    u, v, w = map(int, input().split())
+    G[u - 1][v - 1] = w
+
+def warshall_floyd(n, d):
+    # 空間計算量: O(n^2)
+    # 時間計算量: O(n^3)
+    # よって、n <= 10^2程度が限界
+    for k in range(n):
+        for i in range(n):
+            for j in range(n):
+                d[i][j] = min(d[i][j], d[i][k] + d[k][j])
+    return d
+
+G = warshall_floyd(n, G)
+
+def solve_tsp_dp(n, g):
+    # n: 頂点数
+    # m: 辺数
+    # g: グラフ
+    # https://qiita.com/Ll_e_ki/items/fa475f5bb224ada9be97
+    dp = [[INF] * n for _ in range(2**n)]
+    # どこスタートでも良い
+    for i in range(n):
+        dp[2**i][i] = 0
+    for S in range(2**n):
+        for v in range(n):
+            for u in range(n):
+                if not (S >> u) & 1 and S != 0: # ①
+                    continue
+                if (S >> v) & 1 == 0:
+                    if dp[S][u] + g[u][v] < dp[S | (1 << v)][v]:
+                        dp[S | (1 << v)][v] = dp[S][u] + g[u][v]
+    return min(dp[2**n - 1])
+
+ans = solve_tsp_dp(n, G)
+if ans == INF:
+    print("No")
+else:
+    print(ans)

atcoder/abc/abc301-400/abc338/g.py

@@ -5,7 +5,7 @@ FROM --platform=linux/x86_64 ubuntu:22.04
 
 # ユーザの追加
 ARG UID
-ENV USER_NAME=atcoder_user
+ENV USER_NAME=player
 RUN useradd $USER_NAME -u $UID -m
 
 ARG DEBIAN_FRONTEND=noninteractive