minimal sparse ruler problem 最少尺子刻度问题

一个长度13的尺子,如果在1位置刻点可以量出1和12,13三种刻度.那么至少刻几个点,可以直接量出1-13所有的长度,分别刻在哪几个位置？

注：必须是直接量。即在尺子上能找出一个1-13任意的整数长度。

写了个没什么技术含量的dfs暴力求解。一个可行解是 1, 2, 6, 10。

 1 #include <iostream>
 2 #include <vector>
 3 #include <unordered_map>
 4 using namespace std;
 5 
 6 class Solution {
 7 public:
 8     vector<vector<int>> ruler(int n, vector<int> &minPath) {
 9         dfs(0, n, minPath);
10         return result;
11     }
12     vector<vector<int>> result;
13     vector<int> path;
14     void dfs(int start, int n, vector<int> &minPath) {
15         if (start == n + 1) {
16             if (fullScale(path)) {
17                 result.push_back(path);
18                 if (path.size() < minLen) {
19                     minLen = path.size();
20                     minPath = path;
21                 }
22             }
23             return;
24         }
25         for (int i = start; i <= n; i++) {
26             path.push_back(i);
27             dfs(i + 1, n, minPath);
28             path.pop_back();
29         }
30     }
31     bool fullScale(vector<int> path) {
32         if (path.size() < 4) {
33             return false;
34         }
35         unordered_map<int, int> umap;
36         umap[13]++;
37         umap[0]++;
38         for (int i = 0; i < path.size(); i++) {
39             for (int j = 0; j < i; j++) {
40                 if (path[i] - path[j] < 13) {
41                     umap[path[i] - path[j]]++;
42                     umap[path[j]]++;
43                     umap[path[i]]++;
44                     umap[13 - path[i]]++;
45                     umap[13 - path[j]]++;
46                 }
47                 if (umap.size() >= 14) {
48                     return true;
49                 }
50             }
51         }
52         return false;
53     }
54 private:
55     int minLen = 13;
56 };
57 
58 int main() {
59     int n = 13;
60     Solution solu;
61     vector<int> minPath;
62     vector<vector<int>> res = solu.ruler(n, minPath);
63     for (auto x : minPath) {
64         cout << x << ", ";
65     }
66 }

ref: https://en.wikipedia.org/wiki/Sparse_ruler