init
This commit is contained in:
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144
SourceCollector.py
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144
SourceCollector.py
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import csv
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class TreeNode:
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def __init__(self):
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self.fa = None
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self.children = []
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self.pos = None
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self.val = 0
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self.id = 0
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self.dp = 0
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self.path = []
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self.dp_path = []
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class SourceCollector:
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def __init__(self, filename, maze=None):
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self.filename = filename
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self.maze = maze
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self.start_pos = None
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self.end_pos = None
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self.path = []
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if self.filename:
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self.maze = []
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with open(f"{self.filename}",'r') as f:
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reader = csv.reader(f)
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for row in reader:
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t = []
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for i in row:
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if i.startswith('b') or i.startswith('l'):
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t.append('0')
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else:
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t.append(i)
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self.maze.append(t)
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self.rowNums = len(self.maze)
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self.colNums = len(self.maze[0])
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for i in range(self.rowNums):
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for j in range(self.colNums):
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if self.maze[i][j] =='s':
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self.start_pos = (i,j)
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if self.maze[i][j] =='e':
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self.end_pos = (i,j)
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def dfs_show(self,u):
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if u.id != 0:
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print(f"id: {u.id} , fa:{u.fa.id} , val:{u.val}")
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else:
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print(f"id: {u.id} , val:{u.val}")
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print(u.path)
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for child in u.children:
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self.dfs_show(child)
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def outofmap(self,x,y):
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return x < 0 or y < 0 or x > self.rowNums or y > self.colNums
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def build_a_tree(self):
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sn = TreeNode()
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sn.pos = self.start_pos
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id = 0
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sn.id = id
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sn.val = 0
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sn.path = [sn.pos]
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st = [[False] * self.colNums for _ in range(self.rowNums)]
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qsk = []
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sx, sy = self.start_pos
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st[sx][sy] = True
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qsk.append((self.start_pos[0],self.start_pos[1], sn,[]))
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dx = [-1,0,1,0]
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dy = [0,-1,0,1]
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while qsk:
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x, y, u,path = qsk.pop()
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for _x, _y in zip(dx,dy):
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nx, ny = x + _x, y + _y
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if self.outofmap(nx,ny):
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continue
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if self.maze[nx][ny] == '1' or st[nx][ny]:
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continue
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st[nx][ny] = True
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to = self.maze[nx][ny]
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new_path = path + [(nx,ny)]
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if to.startswith('g') or to.startswith('t'):
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new_node = TreeNode()
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id+=1
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new_node.id = id
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new_node.pos = (nx, ny)
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new_node.fa = u
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num_str = to[1:]
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new_node.path = new_path
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try:
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if to.startswith('g'):
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new_node.val = int(num_str)
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else:
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new_node.val = -int(num_str)
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except ValueError:
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print("wa ! ")
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u.children.append(new_node)
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qsk.append((nx, ny, new_node,[]))
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else:
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qsk.append((nx, ny, u,new_path))
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return sn
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def dfs(self,sn):
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sn.dp = sn.val
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for child in sn.children:
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# print(f"cur : {child.pos} , fa : {child.fa.pos} , childrens : {child.path}")
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self.dfs(child)
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if child.dp > 0:
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sn.dp += child.dp
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sn.dp_path += child.path + child.dp_path + child.path[::-1]
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def find_path(self):
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self.path = []
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sn = self.build_a_tree()
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self.dfs(sn)
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return self.path
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if __name__ == '__main__':
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obj = SourceCollector(filename="maze.csv")
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sn = obj.build_a_tree()
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# obj.dfs_show(sn)
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obj.dfs(sn)
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print(len(sn.dp_path))
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print(sn.pos)
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pre = sn.pos
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for _ in sn.dp_path:
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dx,dy = _[0] - pre[0],_[1]-pre[1]
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if dx > 0:
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print("down")
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elif dx < 0:
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print("up")
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elif dy > 0:
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print("right")
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elif dy < 0:
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print("left")
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pre = _
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136
boss_fight.py
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136
boss_fight.py
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# -*-coding: GBK -*-
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import heapq # 导入堆模块，用于实现优先队列
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from typing import List, Optional, Tuple # 导入类型提示相关模块
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def boss_strategy(coin: int, blood: int) -> Optional[Tuple[List[str], int]]:
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# 使用分支限界法寻找最小回合数击败BOSS的技能序列
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# 参数:
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# coin: 玩家初始金币数
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# blood: BOSS初始血量
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# 返回:
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# 最小回合数的技能序列列表，如果无解则返回None
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start_coin = coin # 记录初始金币数，用于计算消耗金币
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# 初始状态: (回合数, 剩余金币, BOSS血量, 充能, 连续E计数, 技能序列元组)
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start_state = (0, coin, blood, 0, 0, ())
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# 创建优先队列: 元素为 (优先级, 状态)
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# 优先级元组: (回合数, 消耗金币, BOSS血量) - 值越小优先级越高
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heap = []
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heapq.heappush(heap, ((0, 0, blood), start_state)) # 将初始状态加入优先队列
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# 创建访问集合: 用于避免重复状态
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# 状态键: (回合数, 剩余金币, BOSS血量, 充能, min(连续E计数,3))
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visited = set()
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start_ce_compressed = min(0, 3) # 压缩连续E计数状态
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visited.add((0, coin, blood, 0, start_ce_compressed)) # 将初始状态加入已访问集合
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# 设置最大回合数限制，避免无限循环
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max_round = blood * 2 # 每次至少造成1点伤害，最多回合数为血量两倍
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# 主循环: 处理优先队列中的状态
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while heap:
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# 从优先队列中取出优先级最高的状态
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priority, state = heapq.heappop(heap)
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# 解包状态元组
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round_cnt, cur_coin, cur_blood, energy, ce, seq_tuple = state
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# 检查是否已击败BOSS
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if cur_blood <= 0:
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return (list(seq_tuple), cur_coin) # 返回技能序列和剩余金币
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# 超过最大回合数则跳过当前状态
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if round_cnt >= max_round:
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continue
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# 尝试使用技能A
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if cur_coin >= 2: # 检查金币是否足够
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new_coin = cur_coin - 2 # 扣除金币
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new_blood = cur_blood - 1 # 减少BOSS血量
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new_energy = energy + 1 # 增加充能
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new_ce = 0 # 使用非E技能，重置连续E计数
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new_round = round_cnt + 1 # 回合数+1
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new_seq = seq_tuple + ('A',) # 添加技能到序列
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# 压缩连续E计数状态（>=3视为3）
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ce_compressed = min(new_ce, 3)
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# 创建状态键用于去重
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state_key = (new_round, new_coin, new_blood, new_energy, ce_compressed)
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# 检查是否为新状态
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if state_key not in visited:
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visited.add(state_key) # 标记为已访问
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cost = start_coin - new_coin # 计算已消耗金币
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# 创建新状态
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new_state = (new_round, new_coin, new_blood, new_energy, new_ce, new_seq)
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# 将新状态加入优先队列
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heapq.heappush(heap, ((new_round, cost, new_blood), new_state))
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# 尝试使用技能E
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cost_e = 3 # 默认消耗金币
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if ce >= 2: # 检查是否为连续第三次使用E
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cost_e = 2 # 第三次消耗2金币
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if cur_coin >= cost_e: # 检查金币是否足够
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new_coin = cur_coin - cost_e # 扣除金币
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new_blood = cur_blood - 3 # 减少BOSS血量
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new_energy = energy + 1 # 增加充能
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new_ce = ce + 1 # 增加连续E计数
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new_round = round_cnt + 1 # 回合数+1
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new_seq = seq_tuple + ('E',) # 添加技能到序列
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# 压缩连续E计数状态
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ce_compressed = min(new_ce, 3)
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# 创建状态键用于去重
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state_key = (new_round, new_coin, new_blood, new_energy, ce_compressed)
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# 检查是否为新状态
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if state_key not in visited:
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visited.add(state_key) # 标记为已访问
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cost = start_coin - new_coin # 计算已消耗金币
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# 创建新状态
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new_state = (new_round, new_coin, new_blood, new_energy, new_ce, new_seq)
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# 将新状态加入优先队列
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heapq.heappush(heap, ((new_round, cost, new_blood), new_state))
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# 尝试使用技能Q
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if energy >= 3 and cur_coin >= 3: # 检查充能和金币是否足够
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new_coin = cur_coin - 3 # 扣除金币
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new_blood = cur_blood - 8 # 减少BOSS血量
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new_energy = energy - 3 # 消耗充能
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new_ce = 0 # 使用非E技能，重置连续E计数
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new_round = round_cnt + 1 # 回合数+1
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new_seq = seq_tuple + ('Q',) # 添加技能到序列
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# 压缩连续E计数状态
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ce_compressed = min(new_ce, 3)
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# 创建状态键用于去重
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state_key = (new_round, new_coin, new_blood, new_energy, ce_compressed)
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# 检查是否为新状态
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if state_key not in visited:
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visited.add(state_key) # 标记为已访问
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cost = start_coin - new_coin # 计算已消耗金币
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# 创建新状态
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new_state = (new_round, new_coin, new_blood, new_energy, new_ce, new_seq)
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# 将新状态加入优先队列
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heapq.heappush(heap, ((new_round, cost, new_blood), new_state))
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return None # 队列为空，无解
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# 主程序入口
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if __name__ == "__main__":
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# 测试示例: 金币=10, BOSS血量=10
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result = boss_strategy(20, 20)
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if result is None:
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print("无法击败BOSS")
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else:
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skill_sequence, remaining_coins = result
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print("最小回合数技能序列:", skill_sequence) # 预期输出: ['A', 'A', 'A', 'Q']
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print("剩余金币:", remaining_coins) # 预期输出: 1
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30
maze.csv
30
maze.csv
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1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1
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1,0,l30,e,0,1,0,1,0,t17,1,0,g13,1,g23,1
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1,1,1,1,0,1,0,0,0,0,1,0,0,1,0,1
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1,0,0,t14,0,0,0,1,0,0,1,1,0,1,t20,1
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1,0,t17,0,0,1,0,1,0,0,1,0,0,1,0,1
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1,1,0,1,1,1,1,1,1,1,1,0,0,1,0,1
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1,0,0,0,l19,0,0,1,0,0,1,0,0,1,0,1
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1,1,0,1,1,1,1,1,0,0,0,0,0,1,0,1
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1,0,l30,0,0,t18,0,g14,0,0,1,0,0,0,0,1
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1,1,1,1,1,1,1,1,1,1,1,0,1,1,1,1
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1,0,0,g26,0,0,0,0,0,b66,1,0,1,0,0,1
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1,1,0,1,0,1,1,1,0,t20,1,l22,1,0,g27,1
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1,0,0,1,g16,0,0,1,0,0,0,t17,s,0,0,1
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1,1,1,1,1,1,1,1,g12,1,1,0,1,0,1,1
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1,0,0,0,0,0,0,t7,0,0,1,0,1,0,0,1
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1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1
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1,0,0,1,0,e,t7,1,0,0,0,0,0,g19,0,1
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1,0,1,1,0,1,1,1,1,1,1,1,1,1,0,1
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1,0,l25,0,t5,0,0,1,0,0,0,0,t9,0,g27,1
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1,0,1,1,1,1,1,1,1,1,1,0,1,1,1,1
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1,0,1,0,1,0,0,1,0,0,0,0,1,0,0,1
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1,0,1,0,1,0,0,1,1,1,1,0,0,0,0,1
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1,0,0,0,1,0,0,1,0,0,0,0,1,0,0,1
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1,1,1,0,0,0,0,1,1,1,1,0,1,0,1,1
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1,l25,g10,0,1,0,0,0,0,1,0,0,1,0,b73,1
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1,1,1,1,1,t9,1,1,g20,1,0,0,1,0,g12,1
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1,0,1,0,1,0,0,1,0,1,0,0,1,0,0,1
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1,0,0,0,1,0,0,1,0,1,0,0,1,0,0,1
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1,1,1,0,0,0,0,1,t7,1,0,1,1,0,0,1
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1,g12,0,0,1,0,s,1,0,0,t13,0,1,l17,0,1
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1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1
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100
readme.md
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100
readme.md
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# 算法驱动的迷宫探险游戏开发
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## 设计目标
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***\*地图：\****
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一个二维矩阵，采⽤分治法⽣成，⽆孤⽴区域，存在唯⼀通路。
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输⼊：迷宫尺⼨ 。
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输出：迷宫矩阵（可存储为JSON或CSV），包括起点Start（S）、 终点Exit（E）、墙壁（#）、通路（空格）、资源（例如⾦币G）、 陷阱Trap（T）、机关Locker（L）、BOSS（B）。
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墙壁：不可通过，无value的节点
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通路：可通过，无value的节点
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金币：可通过，有value 的节点。BOSS战时的行动需要消耗金币
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陷阱：可通过，但经过时会扣除玩家的金币
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机关：三位密码锁。一般分布在某条路的尽头，必须开启所有机关后，终点大门才会打开
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BOSS：在必经之路上，有value(血量) , 必须打败才能继续前进
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***\*玩家：\****
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有三种攻击方式（初始充能为0）：
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A：消耗2枚金币造成1点伤害并获得1点充能
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E：消耗3枚金币造成3点伤害并获得1点充能，连续第三次使用E技能时少消耗1枚金币
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Q：消耗3枚金币和3点充能，造成8点伤害
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## 任务分解
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### 1. 分治法生成迷宫
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**需求 : **
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输入 : 正整数 n( 100>=n>=7 ) , 生成类型选择(json或者csv)
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输出 : json文件 , csv文件 , 表示地图 , 无孤立区 , 存在唯一通路
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S : 起点
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E : 终点
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\# :墙壁
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blank : 通路
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G : 资源
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T : 陷阱
|
||||
|
||||
L : 机关
|
||||
|
||||
B : Boss
|
||||
|
||||
**递归分割法生成迷宫**
|
||||
|
||||
|
||||
|
||||
|
||||
|
||||
|
||||
|
||||
### 2. 动态规划进行资源收集路径规划
|
||||
|
||||
|
||||
|
||||
|
||||
|
||||
### 3. 贪心算法设计实时资源拾取策略
|
||||
|
||||
|
||||
|
||||
|
||||
|
||||
### 4. 回溯法解迷关卡
|
||||
|
||||
|
||||
|
||||
|
||||
|
||||
### 5. 分支限界设计boss战
|
||||
|
||||
|
||||
|
||||
|
||||
|
||||
Loading…
Reference in New Issue
Block a user