這里的star battle游戲不是指別的(像war frame),就是puzzle team club搞的游戲,在https://www.puzzle-star-battle.com/里面可以找到。
這里要解題的話,不能再像上回那樣用舞蹈表(dancing link)了,因為游戲規則決定了方塊的占用位置不是全部都要用,一個方格可以相鄰1個或2個星星,無法像之前那樣使用精確覆蓋的做法。但是,這里要解題還是很簡單的,約束條件的某一種(行內必須正好有n個星星)可以利用這點搞DFS。
star battle的規則如下:放置一定數量的星星在棋盤,使所有的星星鄰近8格沒有星星,且每行、每列、每個區域正好有指定數量的星星,至於指定數量是多少要看星星★左邊是哪個數字
圖1.1★表示每行每列每塊必須正好有1個星星;3★則是正好有3個
這里就需要初始化每行每列每塊占據的星星數為指定數字(這里的深度優先搜索參數就是行數,操作也是基於單行搜索,所以省略了每行占據星星數)

def init(): with open('starBattleChess1.txt','r') as f: chessStr = f.read() rowStrs = chessStr.split('\n') global rowsize, colsize rowsize = len(rowStrs) colsize = len(rowStrs[0].split(' ')) maxnum = 0 for rowStr in rowStrs: row = [] validrow = [] for j in range(2): occupy[j].append(limit) #0是每列,1是每塊 for j in range(limit): answer.append([]) for colStr in rowStr.split(' '): maxnum = maxnum if int(colStr) < maxnum else int(colStr) row.append(int(colStr)) validrow.append(1) chess.append(row) valid.append(validrow)
深度優先搜索部分(基於單行橫向搜索),答案部分用一個謎面長度*限制大小為長度的數組儲存。這里對方格的占據操作使用類似於舞蹈表的消除(remove)和還原(resume)操作

def dfs(depth, occ, start): #print('depth : ' + str(depth) + ' occ : ' + str(occ) + ' start : ' + str(start)) if depth == rowsize: print(answer) printAnswer() return for i in range(start, colsize-limit+occ+1): if occupy[0][i] <= 0 or occupy[1][chess[depth][i]] <= 0 or valid[depth][i] == 0: continue resumePos = [] occupy[0][i] -= 1 occupy[1][chess[depth][i]] -= 1 for j in range(depth-1, depth+2): for k in range(i-1, i+2): if j < 0 or j > rowsize - 1:continue if k < 0 or k > rowsize - 1:continue if valid[j][k] != 0: valid[j][k] = 0 resumePos.append([j, k]) answer[depth * limit + occ] = [depth, i] if occ >= limit - 1: dfs(depth+1, 0, 0) else: dfs(depth, occ+1, i+1) for resume in resumePos: valid[resume[0]][resume[1]] = 1 occupy[0][i] += 1 occupy[1][chess[depth][i]] += 1
輸出答案:

def printAnswer(): defaultImg = ['+','|'] for _ in range(rowsize): defaultImg[0] += '-+' defaultImg[1] += ' |' imgs = [defaultImg[0]] for i in range(rowsize): imgs.append(defaultImg[1]) imgs.append(defaultImg[0]) for ans in answer: a = ans[0] b = ans[1] imgs[2*a+1] = imgs[2*a+1][:2*b+1] + '*' + imgs[2*a+1][2*b+2:] print(reduce(lambda a,b:a+'\n'+b,imgs))
總的代碼:

from functools import reduce import time chess = [] rowsize = 0 colsize = 0 limit = 2 # 看星星★左邊是哪個數字就填哪個 valid = [] occupy = [[],[]]# 0 is vertical; 1 is block answer = [] # def printAnswer(): defaultImg = ['+','|'] for _ in range(rowsize): defaultImg[0] += '-+' defaultImg[1] += ' |' imgs = [defaultImg[0]] for i in range(rowsize): imgs.append(defaultImg[1]) imgs.append(defaultImg[0]) for ans in answer: a = ans[0] b = ans[1] imgs[2*a+1] = imgs[2*a+1][:2*b+1] + '*' + imgs[2*a+1][2*b+2:] print(reduce(lambda a,b:a+'\n'+b,imgs)) # def dfs(depth, occ, start): #print('depth : ' + str(depth) + ' occ : ' + str(occ) + ' start : ' + str(start)) if depth == rowsize: print(answer) printAnswer() return for i in range(start, colsize-limit+occ+1): if occupy[0][i] <= 0 or occupy[1][chess[depth][i]] <= 0 or valid[depth][i] == 0: continue resumePos = [] occupy[0][i] -= 1 occupy[1][chess[depth][i]] -= 1 for j in range(depth-1, depth+2): for k in range(i-1, i+2): if j < 0 or j > rowsize - 1:continue if k < 0 or k > rowsize - 1:continue if valid[j][k] != 0: valid[j][k] = 0 resumePos.append([j, k]) answer[depth * limit + occ] = [depth, i] if occ >= limit - 1: dfs(depth+1, 0, 0) else: dfs(depth, occ+1, i+1) for resume in resumePos: valid[resume[0]][resume[1]] = 1 occupy[0][i] += 1 occupy[1][chess[depth][i]] += 1 # def init(): with open('starBattleChess1.txt','r') as f: chessStr = f.read() rowStrs = chessStr.split('\n') global rowsize, colsize rowsize = len(rowStrs) colsize = len(rowStrs[0].split(' ')) maxnum = 0 for rowStr in rowStrs: row = [] validrow = [] for j in range(2): occupy[j].append(limit) for j in range(limit): answer.append([]) for colStr in rowStr.split(' '): maxnum = maxnum if int(colStr) < maxnum else int(colStr) row.append(int(colStr)) validrow.append(1) chess.append(row) valid.append(validrow) if __name__ == "__main__": init() start = time.time() dfs(0, 0, 0) end = time.time() print('search time : ' + str(end-start) + 's')
我們在同目錄下創建starBattleChess1.txt文件,並錄入棋盤:

0 1 1 1 1 1 2 2 2 2 0 0 0 3 3 1 4 4 2 2 0 0 0 3 3 3 4 4 2 2 0 5 4 4 4 4 4 4 2 2 0 5 4 6 6 6 6 6 2 2 5 5 5 6 6 6 6 6 6 2 5 5 5 5 5 5 7 7 7 7 5 5 5 5 8 5 7 7 7 7 5 9 9 8 8 8 8 7 7 7 9 9 8 8 8 8 8 7 7 7
執行結果:
[[0, 3], [0, 5], [1, 1], [1, 8], [2, 3], [2, 5], [3, 7], [3, 9], [4, 0], [4, 2], [5, 6], [5, 8], [6, 1], [6, 4], [7, 7], [7, 9], [8, 2], [8, 4], [9, 0], [9, 6]] +-+-+-+-+-+-+-+-+-+-+ | | | |*| |*| | | | | +-+-+-+-+-+-+-+-+-+-+ | |*| | | | | | |*| | +-+-+-+-+-+-+-+-+-+-+ | | | |*| |*| | | | | +-+-+-+-+-+-+-+-+-+-+ | | | | | | | |*| |*| +-+-+-+-+-+-+-+-+-+-+ |*| |*| | | | | | | | +-+-+-+-+-+-+-+-+-+-+ | | | | | | |*| |*| | +-+-+-+-+-+-+-+-+-+-+ | |*| | |*| | | | | | +-+-+-+-+-+-+-+-+-+-+ | | | | | | | |*| |*| +-+-+-+-+-+-+-+-+-+-+ | | |*| |*| | | | | | +-+-+-+-+-+-+-+-+-+-+ |*| | | | | |*| | | | +-+-+-+-+-+-+-+-+-+-+ search time : 0.9020516872406006s
這個算法還有優化空間。既然這個網站的棋盤都是唯一解,那肯定可以找出查到唯一解的方式。這個后面再說。
把這個答案錄入網頁:
圖2.不要吐槽時間為什么這么久,3天前就打開過又退出了
嘿嘿,大功告成!
解這道題時,請拋棄1星棋盤的常規解法,多從多星角度考慮。之前我寫代碼就把最后的答案儲存習慣性按1星寫入,結果答案就只剩一半了。。。