solve problems

Author: tiny656

Data Structures and Algorithms (English)/7-10_Saving James Bond - Easy Version (25).py

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+import math
+
+
+def dist(a, b):
+    return math.hypot(a[0] - b[0], a[1] - b[1])
+
+
+def can_escape(croc, D):
+    """判断鳄鱼 croc=(x,y) 能否一步跳到岸"""
+    x, y = croc
+    return 50 - max(abs(x), abs(y)) <= D
+
+
+n, d = list(map(int, input().split()))
+crocs = []
+for _ in range(n):
+    x, y = list(map(int, input().split()))
+    crocs.append((x, y))
+
+# 直接从岛跳到岸
+if d >= 42.5:  # 50 - 7.5
+    print("Yes")
+
+# 找第一跳可达的鳄鱼
+first = [i for i, (x, y) in enumerate(crocs) if dist((0, 0), (x, y)) <= d + 7.5]
+
+visited = [False] * n
+queue = first[:]
+
+is_escaped = False
+while queue and not is_escaped:
+    i = queue.pop()
+    if visited[i]:
+        continue
+    visited[i] = True
+
+    if can_escape(crocs[i], d):
+        is_escaped = True
+
+    if not is_escaped:
+        for j in range(n):
+            if not visited[j] and dist(crocs[i], crocs[j]) <= d:
+                queue.append(j)
+
+print("Yes" if is_escaped else "No")

Data Structures and Algorithms (English)/7-9_Huffman Codes (30).py

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+import heapq
+
+
+def build_wpl(freq):
+    pq = list(freq.values())
+    heapq.heapify(pq)
+    wpl = 0
+    while len(pq) > 1:
+        a = heapq.heappop(pq)
+        b = heapq.heappop(pq)
+        wpl += a + b
+        heapq.heappush(pq, a + b)
+    return wpl
+
+
+def is_prefix_free(codes):
+    sorted_codes = sorted(codes.values())
+    for i in range(len(sorted_codes) - 1):
+        if sorted_codes[i + 1].startswith(sorted_codes[i]):
+            return False
+    return True
+
+
+n = int(input())
+freq = {}
+data = input().split()
+for i in range(0, len(data), 2):
+    freq[data[i]] = int(data[i + 1])
+
+WPL_min = build_wpl(freq)
+
+m = int(input())
+for _ in range(m):
+    codes = {}
+    ok = True
+    cur_wpl = 0
+    for _ in range(n):
+        c, code = input().split()
+        codes[c] = code
+        cur_wpl += len(code) * freq.get(c, 0)
+
+    # condition 1：all characters are present
+    if len(codes) != n:
+        ok = False
+    # condition 2：WPL is minimum
+    if cur_wpl != WPL_min:
+        ok = False
+    # condition 3：prefix-free
+    if ok and not is_prefix_free(codes):
+        ok = False
+    print("Yes" if ok else "No")