Leetcode 464. Can I Win

最新推荐文章于 2025-05-21 03:30:00 发布

zhengjihao

最新推荐文章于 2025-05-21 03:30:00 发布

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分类专栏： ★算法入门 ------动态规划 -----博弈论文章标签： Leetcode 动态规划状态压缩 map

本文链接：https://blog.youkuaiyun.com/zhengjihao/article/details/74895716

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本文探讨了变种100游戏的玩法，在此游戏中两名玩家轮流从1到15中选择不重复的数字，使总和达到或超过100即获胜。文章通过状态压缩动态规划的方法，分析了在给定最大可选数字和目标总和的情况下，先手玩家是否有必胜策略。

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In the “100 game,” two players take turns adding, to a running total, any integer from 1..10. The player who first causes the running total to reach or exceed 100 wins.

What if we change the game so that players cannot re-use integers?

For example, two players might take turns drawing from a common pool of numbers of 1..15 without replacement until they reach a total >= 100.

Given an integer maxChoosableInteger and another integer desiredTotal, determine if the first player to move can force a win, assuming both players play optimally.

You can always assume that maxChoosableInteger will not be larger than 20 and desiredTotal will not be larger than 300.

Example

Input:
maxChoosableInteger = 10
desiredTotal = 11

Output:
false

Explanation:
No matter which integer the first player choose, the first player will lose.
The first player can choose an integer from 1 up to 10.
If the first player choose 1, the second player can only choose integers from 2 up to 10.
The second player will win by choosing 10 and get a total = 11, which is >= desiredTotal.
Same with other integers chosen by the first player, the second player will always win.
原文链接
思路分析：
状态压缩的动态规划算法同时也是带记忆化的dfs算法。数字选取范围最大为不超过20，我们可以用整数的每一位代表一个数字是否被选择，这样就可以用一个整数代表所有数被选择的状态。注意到如果用数组表示，范围太大，所以我们用map表示。
首先判断两种边界情况：第一次选择就成功和不可能成功

代码如下：

class Solution {
private:
    int maxn;
    map<int,bool> mp;
public:
    bool canIWin(int maxChoosableInteger, int desiredTotal) {
        maxn = maxChoosableInteger;
        if(maxn >= desiredTotal) return true;
        if((1 + maxn) * maxn / 2 < desiredTotal) return false;
        return canWin(desiredTotal,0);
    }
    bool canWin(int target, int visited) {
        if(mp.find(visited) != mp.end()) return mp[visited];
        for(int i = 1; i <= maxn; i++) {
            int mask = (1 << (i-1));
            if((mask & visited) == 0 && (i >= target || canWin(target-i, mask | visited) == false)) {
                mp[visited] = true;
                return true;
            }
        }
        mp[visited] = false;
        return false;
    }
};