2020阿里笔试（扑克牌）

题目描述

有一叠扑克牌，每张牌介于1和10之间

有四种出牌方法：
单出1张
出2张对子
出五张顺子，如12345
出三连对子，如112233
给10个数，表示1-10每种牌有几张，问最少要多少次能出完

思路

记忆化搜索+剪枝

def dfs(state):
    if tuple(state) in memo:
        return memo[tuple(state)]
    if sum(state) == 0:
        return 0
    else:
        res = float('inf')
        for i in range(10):
            # 出顺子
            if i < 6 and state[i] > 0 and state[i + 1] > 0 and state[i + 2] > 0 and state[i + 3] > 0 and state[
                i + 4] > 0:
                state[i] -= 1
                state[i + 1] -= 1
                state[i + 2] -= 1
                state[i + 3] -= 1
                state[i + 4] -= 1
                res = min(res, dfs(state) + 1)
                state[i] += 1
                state[i + 1] += 1
                state[i + 2] += 1
                state[i + 3] += 1
                state[i + 4] += 1
                memo[tuple(state)] = res
            # 出三连对子
            if i < 8 and state[i] > 1 and state[i + 1] > 1 and state[i + 2] > 1:
                state[i] -= 2
                state[i + 1] -= 2
                state[i + 2] -= 2
                res = min(res, dfs(state) + 1)
                state[i] += 2
                state[i + 1] += 2
                state[i + 2] += 2
                memo[tuple(state)] = res
            # 出2张对子
            if res == float('inf') and state[i] > 1:
                    state[i] -= 2
                    res = min(res, dfs(state) + 1)
                    state[i] += 2
                    memo[tuple(state)] = res
            # 单出一张
            if res == float('inf') and state[i] > 0:
                state[i] -= 1
                res = min(res, dfs(state)+1)
                state[i] += 1
                memo[tuple(state)] = res
        return res
memo = {}
# state = list(map(int, input().split()))
state = [4,4,4,4,4,4,4,4,3,3]
print(dfs(state))