易错辨析：七级四大翻车现场

# 易错辨析：七级四大翻车现场（错例 vs 正确写法）
import math
from collections import deque

print("== 坑 1：二维列表的别名陷阱 ==")
bad = [[0] * 3] * 3          # 错例：三行其实是同一行
bad[0][1] = 1
print("错例改一格后：", bad)
good = [[0] * 3 for _ in range(3)]
good[0][1] = 1
print("正确改一格后：", good)

print("== 坑 2：BFS 出队才标记 ==")
adj = [[1, 2], [0, 2], [0, 1]]   # 三角形 0-1-2
# 错例：等出队才标记，2 号会被重复入队
q = deque([0])
vis = [False] * 3
push = [1, 0, 0]
while q:
    u = q.popleft()
    if vis[u]:
        continue
    vis[u] = True
    for v in adj[u]:
        if not vis[v]:
            q.append(v)
            push[v] += 1
print("错例各点入队次数：", push)
# 正确：入队那一刻就标记
q = deque([0])
vis = [False] * 3
vis[0] = True
push = [1, 0, 0]
while q:
    u = q.popleft()
    for v in adj[u]:
        if not vis[v]:
            vis[v] = True
            q.append(v)
            push[v] += 1
print("正确各点入队次数：", push)

print("== 坑 3：背包容量正序枚举 ==")
w, v, W = 2, 10, 4   # 只有一件物品：重 2、价值 10，容量 4
dp_bad = [0] * (W + 1)
for j in range(w, W + 1):              # 错例：正序，同一件被反复装
    dp_bad[j] = max(dp_bad[j], dp_bad[j - w] + v)
dp_good = [0] * (W + 1)
for j in range(W, w - 1, -1):          # 正确：倒序，每件只用一次
    dp_good[j] = max(dp_good[j], dp_good[j - w] + v)
print("错例答案：", dp_bad[W], " 正确答案：", dp_good[W])

print("== 坑 4：sin 的角度与弧度 ==")
print("错例 math.sin(30) =", round(math.sin(30), 4))
print("正确 math.sin(math.radians(30)) =", round(math.sin(math.radians(30)), 4))