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。

 #include <iostream>
 #include <vector>
 #include <unordered_map>
 using namespace std;

 class Solution {
 public:
     vector<vector<int>> ruler(int n, vector<int> &minPath) {
         dfs(, n, minPath);
         return result;
     }
     vector<vector<int>> result;
     vector<int> path;
     void dfs(int start, int n, vector<int> &minPath) {
         if (start == n + ) {
             if (fullScale(path)) {
                 result.push_back(path);
                 if (path.size() < minLen) {
                     minLen = path.size();
                     minPath = path;
                 }
             }
             return;
         }
         for (int i = start; i <= n; i++) {
             path.push_back(i);
             dfs(i + , n, minPath);
             path.pop_back();
         }
     }
     bool fullScale(vector<int> path) {
         if (path.size() < ) {
             return false;
         }
         unordered_map<int, int> umap;
         umap[]++;
         umap[]++;
         for (int i = ; i < path.size(); i++) {
             for (int j = ; j < i; j++) {
                 if (path[i] - path[j] < ) {
                     umap[path[i] - path[j]]++;
                     umap[path[j]]++;
                     umap[path[i]]++;
                     umap[ - path[i]]++;
                     umap[ - path[j]]++;
                 }
                 if (umap.size() >= ) {
                     return true;
                 }
             }
         }
         return false;
     }
 private:
     int minLen = ;
 };

 int main() {
     int n = ;
     Solution solu;
     vector<int> minPath;
     vector<vector<int>> res = solu.ruler(n, minPath);
     for (auto x : minPath) {
         cout << x << ", ";
     }
 }

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