华为OD机试 - 王者荣耀匹配机制 - 回溯（Java 2024 D卷 200分）

题目描述

部门准备举办一场王者荣耀表演赛，有10名游戏爱好者参与，分为两队，每队5人。每位参与者都有一个评分，代表他的游戏水平。为了表演赛尽可能精彩，我们需要将10名参赛者分为实力尽量相近的两队。一队的实力可以表示为这一队5名队员的评分总和。现在给定10名参与者的游戏水平评分，请根据上述要求分队，最后输出这两组的实力差绝对值的最小值。

示例 1:
输入: [5, 1, 8, 3, 4, 6, 7, 10, 9, 2]
输出: 1
解释: 分组为 (1, 3, 5, 8, 10) 和 (2, 4, 6, 7, 9)，两组实力差为 1。

示例 2:
输入: [3, 3, 3, 3, 3, 3, 3, 3, 3, 9]
输出: 6
解释: 分组为 (3, 3, 3, 3, 9) 和 (3, 3, 3, 3, 3)，两组实力差为 6。

解题思路

问题分析：
- 需要将10个评分分成两组，每组5人，使得两组的总评分差最小。
- 直接暴力搜索所有可能的组合会生成 ( C_{10}^5 = 252 ) 种组合，对于给定的输入规模是可接受的。
算法选择：
- 使用回溯法生成所有可能的5人组合，计算每组的总评分，并记录最小的差值。
步骤详解：
- 计算所有评分的总和 total。
- 使用回溯法生成所有5人组合的总评分，存入列表 sumList。
- 遍历 sumList，计算每组总评分与 total / 2 的差值，更新最小差值 minDiff。

代码实现

Java

import java.util.Scanner;public class 王者荣耀匹配机制 {private static int min = Integer.MAX_VALUE;private static int totalSum = 0;private static final int ARR_LENGTH = 10;private static final int SINGLE_GROUP_LENGTH = 5;private static int[] arr = new int[ARR_LENGTH];public static void main(String[] args) {Scanner scanner = new Scanner(System.in);for (int i = 0; i < arr.length; i++) {arr[i] = scanner.nextInt();totalSum += arr[i];}int partSum = 0;backtrace(0, 0, partSum);System.out.println(min);}private static void backtrace(int start, int count, int partSum) {if (count == SINGLE_GROUP_LENGTH) {int currentMin = Math.abs(2 * partSum - totalSum);min = Math.min(min, currentMin);return;}for (int i = start; i < ARR_LENGTH; i++) {backtrace(start + 1, count + 1, partSum + arr[i]);}}
}

import java.util.ArrayList;
import java.util.List;
import java.util.Scanner;public class Main {private static List<Integer> sumList = new ArrayList<>();public static void main(String[] args) {Scanner scanner = new Scanner(System.in);int[] nums = new int[10];for (int i = 0; i < 10; i++) {nums[i] = scanner.nextInt();}int total = 0;for (int num : nums) {total += num;}backtrack(nums, 0, 0, 0);int minDiff = Integer.MAX_VALUE;for (int sum : sumList) {int diff = Math.abs(total - 2 * sum);if (diff < minDiff) {minDiff = diff;}}System.out.println(minDiff);}private static void backtrack(int[] nums, int start, int count, int currentSum) {if (count == 5) {sumList.add(currentSum);return;}for (int i = start; i < nums.length; i++) {backtrack(nums, i + 1, count + 1, currentSum + nums[i]);}}
}

Python

def main():nums = list(map(int, input().split()))total = sum(nums)sum_list = []def backtrack(start, count, current_sum):if count == 5:sum_list.append(current_sum)returnfor i in range(start, 10):backtrack(i + 1, count + 1, current_sum + nums[i])backtrack(0, 0, 0)min_diff = float('inf')for s in sum_list:diff = abs(total - 2 * s)if diff < min_diff:min_diff = diffprint(min_diff)if __name__ == "__main__":main()

C++

#include <iostream>
#include <vector>
#include <climits>
using namespace std;vector<int> sumList;void backtrack(int nums[], int start, int count, int currentSum) {if (count == 5) {sumList.push_back(currentSum);return;}for (int i = start; i < 10; i++) {backtrack(nums, i + 1, count + 1, currentSum + nums[i]);}
}int main() {int nums[10];for (int i = 0; i < 10; i++) {cin >> nums[i];}int total = 0;for (int num : nums) {total += num;}backtrack(nums, 0, 0, 0);int minDiff = INT_MAX;for (int sum : sumList) {int diff = abs(total - 2 * sum);if (diff < minDiff) {minDiff = diff;}}cout << minDiff << endl;return 0;
}

JavaScript

function main() {const nums = prompt("Enter 10 numbers separated by spaces").split(' ').map(Number);const total = nums.reduce((a, b) => a + b, 0);const sumList = [];function backtrack(start, count, currentSum) {if (count === 5) {sumList.push(currentSum);return;}for (let i = start; i < 10; i++) {backtrack(i + 1, count + 1, currentSum + nums[i]);}}backtrack(0, 0, 0);let minDiff = Infinity;for (const sum of sumList) {const diff = Math.abs(total - 2 * sum);if (diff < minDiff) {minDiff = diff;}}console.log(minDiff);
}main();

复杂度分析

时间复杂度：
- 生成所有组合的时间复杂度为 ( O(C_{10}^5) = O(252) )，遍历 sumList 的时间复杂度为 ( O(252) )，因此总时间复杂度为 ( O(1) )（因为输入规模固定为10）。
空间复杂度：
- 存储所有组合的总评分需要 ( O(C_{10}^5) = O(252) ) 的空间，因此空间复杂度为 ( O(1) )。