104 lines
3.3 KiB
GDScript
104 lines
3.3 KiB
GDScript
# VERTICAL SLICE - NOT FOR PRODUCTION
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# Validation Question: 玩家能否无指导在一场完整战斗中经历「开战→挖掘规划→轮末结算逐行兑现→胜/负」并逐行归因（P2）
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# Date: 2026-09-10
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# 棋盘产物数据契约（排雷玩法消费；生成后不可变——排雷玩法持快照即持活引用，生成器不再触碰）。
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# 实现 GDD：design/archive/gdd-2026-09-14/a1-board-generation.md（系统 ID A1·棋盘生成——registry 追溯锚，命名规范见 CLAUDE.md）
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class_name BoardData
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extends RefCounted
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var width: int = 0
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var height: int = 0
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var spawn_center := Vector2i.ZERO
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var mines := {} # Vector2i → true
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var digits := {} # Vector2i → int（安全格恒有值；数字恒真 = 8 邻雷数）
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var variants := {} # Vector2i → Array[String]（生成期快照，抗表重载）
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var board_type := "" # 图类型标签（词表 v1.0：dungeon/underdark；玩家侧显示名「地图类型」）
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var level_index: int = 1
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var node_depth: int = 0
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var metrics := {} # 生成期指标集（键清单单点定义在 board_generator._finalize）
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var illegal_product := false # 非法产物标记（接受最不坏出口——零合法板诚实降级）
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var degraded_product := false # 降级产物标记（接受最优出口——有合法板但无一达标；与 illegal_product 对称）
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func in_bounds(c: Vector2i) -> bool:
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return c.x >= 0 and c.y >= 0 and c.x < width and c.y < height
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func index_of(c: Vector2i) -> int:
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return c.y * width + c.x
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func is_mine(c: Vector2i) -> bool:
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return mines.has(c)
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func is_safe(c: Vector2i) -> bool:
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return in_bounds(c) and not mines.has(c)
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func digit_of(c: Vector2i) -> int:
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return digits.get(c, -1)
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func variant_ids(c: Vector2i) -> Array:
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return variants.get(c, [])
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func has_variant(c: Vector2i, vid: String) -> bool:
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return variant_ids(c).has(vid)
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func neighbors8(c: Vector2i) -> Array:
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var out: Array[Vector2i] = []
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for dy in range(-1, 2):
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for dx in range(-1, 2):
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if dx == 0 and dy == 0:
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continue
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var n := Vector2i(c.x + dx, c.y + dy)
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if in_bounds(n):
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out.append(n)
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return out # 行主序（规范序的邻接枚举基序）
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func spawn_cells() -> Array:
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var out: Array[Vector2i] = []
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for dy in range(-1, 2):
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for dx in range(-1, 2):
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var c := Vector2i(spawn_center.x + dx, spawn_center.y + dy)
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if in_bounds(c):
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out.append(c)
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return out
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func all_cells() -> Array:
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var out: Array[Vector2i] = []
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for y in height:
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for x in width:
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out.append(Vector2i(x, y))
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return out
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func safe_cells() -> Array:
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var out: Array[Vector2i] = []
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for c in all_cells():
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if is_safe(c):
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out.append(c)
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return out
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# R₀ = 出生 3×3 ∪ （出生 3×3 经 0 格的 flood 闭包）——与 排雷玩法 F3 逐字同构（棋盘生成 F1 求解器初始态）。
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func compute_r0() -> Array:
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var b3 := spawn_cells()
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var r0 := {}
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for c in b3:
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r0[c] = true
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var queue: Array = b3.duplicate()
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while not queue.is_empty():
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var c: Vector2i = queue.pop_front()
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if is_safe(c) and digit_of(c) == 0: # 变异死分支退役（2026-09-18 用户拍板删——与 排雷玩法 展开实现口径归一；变异定稿批按正式语义重接）
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for n in neighbors8(c):
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if is_safe(n) and not r0.has(n):
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r0[n] = true
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queue.append(n)
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return r0.keys()
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