ICode9

精准搜索请尝试: 精确搜索
首页 > 其他分享> 文章详细

[Google] LeetCode 2172 Maximum AND Sum of Array 状态压缩DP

2022-09-06 03:01:16  阅读:47  来源: 互联网

标签:slot Google numSlots int Sum nums Maximum state sum


You are given an integer array nums of length n and an integer numSlots such that 2 * numSlots >= n. There are numSlots slots numbered from 1 to numSlots.

You have to place all n integers into the slots such that each slot contains at most two numbers. The AND sum of a given placement is the sum of the bitwise AND of every number with its respective slot number.

For example, the AND sum of placing the numbers [1, 3] into slot 1 and [4, 6] into slot 2 is equal to (1 AND 1) + (3 AND 1) + (4 AND 2) + (6 AND 2) = 1 + 1 + 0 + 2 = 4.
Return the maximum possible AND sum of nums given numSlots slots.

Solution

数据范围很小,这就暗示了这题使用状态压缩。我们用二进制数来表示这些 \(slot\) 的使用情况,如果 \(i\) 位置为 \(1\),说明还未被使用。接着对于每一个 \(num[i]\) 我们枚举每个 \(slot\)。

点击查看代码
class Solution {
private:
    
    vector<vector<int>> dp;
    
    int dfs(int ft, int se, vector<int>& nums, int idx, int& numSlots){
        if(idx==nums.size()) return 0;
        if(dp[ft][se]!=-1) return dp[ft][se];
        int res=0;
        int ans=0;
        for(int i=0;i<numSlots;i++){
            int curslot = 1<<i;
            if(se&curslot){
                res = (i+1)&nums[idx];
                if(ft&curslot)res+=dfs(ft^curslot,se,nums,idx+1,numSlots);
 else{res+=dfs(ft,se^curslot, nums,idx+1,numSlots);} 
            }
            
            ans=max(ans,res);
        }
        return dp[ft][se]=ans;
    }
    
public:
    int maximumANDSum(vector<int>& nums, int numSlots) {
        int state = (1<<numSlots) - 1;
        dp.resize(state+1,vector<int>(state+1,-1));
        int state2 = state;
        return dfs(state, state2, nums, 0, numSlots);
    }
};

标签:slot,Google,numSlots,int,Sum,nums,Maximum,state,sum
来源: https://www.cnblogs.com/xinyu04/p/16660309.html

本站声明: 1. iCode9 技术分享网(下文简称本站)提供的所有内容,仅供技术学习、探讨和分享;
2. 关于本站的所有留言、评论、转载及引用,纯属内容发起人的个人观点,与本站观点和立场无关;
3. 关于本站的所有言论和文字,纯属内容发起人的个人观点,与本站观点和立场无关;
4. 本站文章均是网友提供,不完全保证技术分享内容的完整性、准确性、时效性、风险性和版权归属;如您发现该文章侵犯了您的权益,可联系我们第一时间进行删除;
5. 本站为非盈利性的个人网站,所有内容不会用来进行牟利,也不会利用任何形式的广告来间接获益,纯粹是为了广大技术爱好者提供技术内容和技术思想的分享性交流网站。

专注分享技术,共同学习,共同进步。侵权联系[81616952@qq.com]

Copyright (C)ICode9.com, All Rights Reserved.

ICode9版权所有