华为OD机试-MELON的难题-DFS（JAVA 2025A卷）

题意是从N快雨花石中找出最少拿出雨花石的块数，使得雨花石可以均分，直接使用dfs解决此类组合问题

package com.example.demo.bean;


import java.util.Arrays;
import java.util.LinkedList;
import java.util.Scanner;

public class YuHuaStone {
    public static void main(String[] args) {
        Scanner sc = new Scanner(System.in);
        int n = Integer.parseInt(sc.nextLine());
        int[] nums = new int[n];
        for (int i = 0; i < n; i++) {
            nums[i] = sc.nextInt();
        }
        System.out.println(yuHuaStone(nums));
    }

    private static int res = Integer.MAX_VALUE;

    private static int yuHuaStone(int[] nums) {
        int sum = Arrays.stream(nums).sum();
        if (sum % 2 != 0) return -1;
        int[] used = new int[nums.length];
        LinkedList<Integer> path = new LinkedList<>();
        Arrays.sort(nums);
        dfs(nums, nums.length - 1, path, used, sum / 2);
        return res == Integer.MAX_VALUE ? -1 : res;
    }

    private static void dfs(int[] nums, int start, LinkedList<Integer> path, int[] used, int target) {
        if (target == 0) {
            res = Math.min(res, path.size());
            return;
        }
        for (int i = start; i >= 0; i--) {
            // 雨花石累加和>target，或者累加的雨花石数目>已算出的最小雨花石数目，剪枝
            if (target <= 0 || path.size() >= res) break;
            // 同一层内，重复重量的雨花石直接剪枝
            if (i < nums.length - 1 && nums[i + 1] == nums[i] && used[i + 1] == 0) continue;
            path.addLast(nums[i]);
            used[i] = 1;
            dfs(nums, i - 1, path, used, target - nums[i]);
            path.removeLast();
            used[i] = 0;
        }
    }
}