Leetcode cooking

Welcome all brave people to Likou city , There are cooking pots in the city for the brave to make food , Recover for yourself .

The brave backpack has a total number of  0 ~ 4  Five ingredients , among  materials[j]  It means the first one  j  The number of ingredients . These ingredients can be used to make several dishes ,cookbooks[i][j]  Means to make the first  i  This kind of food needs a second  j  The number of ingredients , and  attribute[i] = [x,y]  It means the first one  i  The delicacy of the dish  x  And satiety  y.

The feeling of fullness is not less than  limit  Under the circumstances , Please return to the greatest delicacy available to the brave . If you can't meet the requirements of satiety , Then return to  -1.

Be careful ：

• Each dish can only be made once .

Example 1：

Input ：materials = [3,2,4,1,2]
cookbooks = [[1,1,0,1,2],[2,1,4,0,0],[3,2,4,1,0]]
attribute = [[3,2],[2,4],[7,6]]
limit = 5

Output ：7

explain ：
The quantity of ingredients can meet the following two schemes ：
Scheme 1 ： Make food 0 And cooking 1, You can get a sense of fullness 2+4、 Delicacy 3+2
Option two ： Only cooking 2, But the feeling of fullness is 6、 The delicacy is 7
Therefore, in meeting the requirements of satiety , Get the highest taste 7

Example 2：

Input ：materials = [10,10,10,10,10]
cookbooks = [[1,1,1,1,1],[3,3,3,3,3],[10,10,10,10,10]]
attribute = [[5,5],[6,6],[10,10]]
limit = 1

Output ：11

explain ： By making food 0 and 1, It can satisfy the feeling of fullness , And get the highest taste 11

Tips ：

• materials.length == 5
• 1 <= cookbooks.length == attribute.length <= 8
• cookbooks[i].length == 5
• attribute[i].length == 2
• 0 <= materials[i], cookbooks[i][j], attribute[i][j] <= 20
• 1 <= limit <= 100

C++

class Solution {
public:
    void dfs(vector<int>& materials, vector<vector<int>>& cookbooks, vector<vector<int>>& attribute, int k) {
        if(k==cookbooks.size()) {
            return;
        }
        for(int i=k;i<cookbooks.size();i++) {
            satiety+=attribute[i][1];
            yummy+=attribute[i][0];
            int flag=0;
            for(int j=0;j<5;j++) {
                used[j]+=cookbooks[i][j];
                if(used[j]>materials[j]) {
                    flag=1;
                }
            }
            if (flag == 0) {
                if (satiety >= limit) {
                    isFull=1;
                    res = max(res, yummy);
                }
                dfs(materials, cookbooks, attribute, i + 1);
            }
            for (int j = 0; j < 5; j++) {
                used[j] -= cookbooks[i][j];
            }
            satiety-=attribute[i][1];
            yummy-=attribute[i][0];
        }
    }

    int perfectMenu(vector<int>& materials, vector<vector<int>>& cookbooks, vector<vector<int>>& attribute, int limit) {
        used.resize(5);
        this->limit=limit;
        dfs(materials,cookbooks,attribute,0);
        if(isFull==0) {
            return -1;
        }
        return res;
    }
private:
    int satiety=0;
    int yummy=0;
    int isFull=0;
    std::vector<int> used;
    int limit;
    int res=0;
};