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

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);
    }
};