Order Problem
Similar Problems:
Description
There is now an order with demand for n items, and the demand for the i-th item is order[i]. The factory has m production modes. Each production mode is shaped like [p[1],p[2],...p[n]], that is, produce p[1] first items, p[2] second items... You can use multiple production modes. Please tell me how many items do not meet the demand at least in the case of not exceeding the demand of any kind of items?
- 1 <= n, m <= 7
- 1 <= order[i] <= 10
- 0 <= pattern[i][j] <= 10
Example Given order=[2,3,1], pattern=[[2,2,0],[0,1,1],[1,1,0]] , return 0. Explanation: Use [0,1,1] once, [1,1,0] twice, remaining [0,0,0].
Given order=[2,3,1], pattern=[[2,2,0]] , return 2. Explanation: Use [2,2,0] once, remaining [0,1,1].
Github: code.dennyzhang.com
Credits To: lintcode.com
Leave me comments, if you have better ways to solve.
- Solution: BFS
#knapsack problem, Ones and Zeroes
// Blog link: https://code.dennyzhang.com/order-problem // Basic Ideas: BFS // Complexity: type Item struct { sum int pattern []int } func getMinRemaining (order []int, pattern [][]int) int { total_sum := 0 for _, v := range order { total_sum += v } sum_list := make([]int, len(pattern)) res := total_sum queue := []Item{} for i, p := range pattern { could_match := true for j, v:= range p { sum_list[i] += v if v > order[j] { could_match = false } } if sum_list[i]!=0 && could_match { queue = append(queue, Item{sum_list[i], p}) if res > total_sum - sum_list[i] { res = total_sum - sum_list[i] } } } for len(queue) != 0 { l := []Item{} for _, item := range queue { for i, p := range pattern { j := 0 newPattern := []int{} diff := total_sum - (sum_list[i] + item.sum) if diff >= 0 { for j<len(order) && p[j]+item.pattern[j]<=order[j] { newPattern = append(newPattern, p[j]+item.pattern[j]) j++ } } if j == len(order) { if diff ==0 { return 0 } l = append(l, Item{sum_list[i] + item.sum, newPattern}) if (diff < res) { res = diff } } } } queue = []Item{} for _, item := range l { queue = append(queue, item) } } return res }
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