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第一个版本对动态规划的理解
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# 问题有大量的重复问题,比如求feibolaqie(5) = feibolaqie(4)+feibolaqie(3), # 所以有重复问题,通过缓存优化,把以前求过的问题做缓存 # def feibolaqie(n): # if n ==1: # return 1 # elif n==2: # return 1 # else: # return feibolaqie(n-1)+feibolaqie(n-2) # print(feibolaqie(7)) # 题目有个机器人,在步长为n的数组中,从开始start位置,走K步到aim位置 # def ways(n, start, aim, k): # return process(start, aim, k, n) # # # def process(cur, aim, rest, n): # if rest == 0: # return 1 if cur == aim else 0 # if cur == 1: # return process(2, aim, rest - 1, n) # elif cur == n: # return process(n - 1, aim, rest - 1, n) # else: # return process(cur - 1, aim, rest - 1, n) + process(cur + 1, aim, rest - 1, n) # 使用缓存对递归经行问题优化,傻缓存法 # def ways1(n, start, aim, k): # # 定义一个[n+1][k+1]的二维数组初始话为-1表示数据没有被复制,这里要用N因为可以走回去 # dp = [[-1 for i in range(k+1)] for j in range(n+1)] # return process1(start, aim, k, n, dp) # # # def process1(cur, aim, rest, n, dp): # if dp[cur][rest] != -1: # return dp[cur][rest] # ans = 0 # if rest == 0: # ans = 1 if cur == aim else 0 # elif cur == 1: # ans = process1(2, aim, rest - 1, n, dp) # elif cur == n: # ans = process1(n - 1, aim, rest - 1, n, dp) # else: # ans = process1(cur - 1, aim, rest - 1, n, dp) + process1(cur + 1, aim, rest - 1, n, dp) # dp[cur][rest] = ans # return ans # 动态规划解决问题 def ways2(n, start, aim, k): # 定义一个[n+1][k+1]的二维数组初始话为-1表示数据没有被复制,这里要用N因为可以走回去,全部定义为0因为rest为0的时候只有目标值为1 dp = [[0 for i in range(k + 1)] for j in range(n + 1)] # 只需要填充退出条件为rest = 0,因为rest为0的时候只有目标值为1 dp[aim][0] = 1 # 填充dp for rest in range(1, k + 1): dp[1][rest] = dp[2][rest - 1] for cur in range(2, n): dp[cur][rest] = dp[cur - 1][rest - 1] + dp[cur + 1][rest - 1] dp[n][rest] = dp[n - 1][rest - 1] return dp[start][k] print(ways2(5, 2, 4, 6))
题目给定一个整数arr,代表不同的数字>0,玩家A和玩家B依次拿走数组的最左或最右,AB都决定聪明,返回获胜者分数
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# 题目给定一个整数arr,代表不同的数字>0,玩家A和玩家B依次拿走数组的最左或最右,AB都决定聪明,返回获胜者分数 def win1(arr): return max(f(arr, 0, len(arr) - 1), g(arr, 0, len(arr) - 1)) # 先手函数,我要选择当前值和后手值相加最大的 def f(arr, l, r): if l == r: return arr[l] p1 = arr[l] + g(arr, l + 1, r) p2 = arr[r] + g(arr, l, r - 1) return max(p1, p2) # 后手函数,只能让先手选择最小的值,因为我们没得选 def g(arr, l, r): if l == r: return 0 p1 = f(arr, l + 1, r) p2 = f(arr, l, r - 1) return min(p1, p2) # 缓存法 def win2(arr): n = len(arr) fcash = [[-1 for _ in range(n)] for _ in range(n)] gcash = [[-1 for _ in range(n)] for _ in range(n)] return max(f2(arr, 0, n - 1, fcash, gcash), g2(arr, 0, n - 1, fcash, gcash)) # 先手函数,我要选择当前值和后手值相加最大的 def f2(arr, l, r, fcash, gcash): if fcash[l][r] != -1: return fcash[l][r] if l == r: ans = arr[l] else: p1 = arr[l] + g2(arr, l + 1, r, fcash, gcash) p2 = arr[r] + g2(arr, l, r - 1, fcash, gcash) ans = max(p1, p2) fcash[l][r] = ans return ans # 后手函数,只能让先手选择最小的值,因为我们没得选 def g2(arr, l, r, fcash, gcash): if gcash[l][r] != -1: return gcash[l][r] if l == r: ans = 0 else: p1 = f2(arr, l + 1, r, fcash, gcash) p2 = f2(arr, l, r - 1, fcash, gcash) ans = min(p1, p2) gcash[l][r] = ans return ans # 动态规划 def win3(arr): n = len(arr) fcash = [[-1 for _ in range(n)] for _ in range(n)] gcash = [[0 for _ in range(n)] for _ in range(n)] for l in range(n): fcash[l][l] = arr[l] for cow in range(1, n): l = 0 r = cow while r < n: fcash[l][r] = max(arr[l] + gcash[l + 1][r], arr[r] + gcash[l][r - 1]) gcash[l][r] = min(fcash[l + 1][r], fcash[l][r - 1]) l += 1 r += 1 return max(fcash[0][n - 1], gcash[0][n - 1]) print(win1([5, 7, 4, 5, 8, 1, 6, 0, 3, 4, 6, 1, 7])) print(win2([5, 7, 4, 5, 8, 1, 6, 0, 3, 4, 6, 1, 7])) print(win3([5, 7, 4, 5, 8, 1, 6, 0, 3, 4, 6, 1, 7]))