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(0, n, minPath);
        return result;
    }
    vector<vector<int>> result;
    vector<int> path;
    void dfs(int start, int n, vector<int> &minPath) {
        if (start == n + 1) {
            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 + 1, n, minPath);
            path.pop_back();
        }
    }
    bool fullScale(vector<int> path) {
        if (path.size() < 4) {
            return false;
        }
        unordered_map<int, int> umap;
        umap[13]++;
        umap[0]++;
        for (int i = 0; i < path.size(); i++) {
            for (int j = 0; j < i; j++) {
                if (path[i] - path[j] < 13) {
                    umap[path[i] - path[j]]++;
                    umap[path[j]]++;
                    umap[path[i]]++;
                    umap[13 - path[i]]++;
                    umap[13 - path[j]]++;
                }
                if (umap.size() >= 14) {
                    return true;
                }
            }
        }
        return false;
    }
private:
    int minLen = 13;
};

int main() {
    int n = 13;
    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