华为OD机试:项目排期
题目描述
项目组共有 N 个开发人员,项目经理接到了 M 个独立的需求,每个需求的工作量不同,且每个需求只能由一个开发人员独立完成,不能多人合作。
假定各个需求直接无任何先后依赖关系,请设计算法帮助项目经理进行工作安排,使整个项目能用最少的时间交付。
输入描述
第一行输入为 M 个需求的工作量,单位为天,用逗号隔开。
例如:
X1 X2 X3 ... Xm
表示共有 M 个需求,每个需求的工作量分别为X1天,X2天,...,Xm天。其中:
0 < M < 30
0 < Xm < 200
第二行输入为项目组人员数量N
例如:
5
表示共有5名员工,其中 0 < N < 10
输出描述
最快完成所有工作的天数
例如:
25
表示最短需要25天完成所有工作
用例
题目解析
1.首先将输入的需求工作量转换为一个列表,例如输入为"6 2 7 7 9 3 2 1 3 11 4",则需求工作量列表为[6, 2, 7, 7, 9, 3, 2, 1, 3, 11, 4]。
2.对需求工作量列表进行排序,从小到大排列,以便后续分配给员工。
3.初始化一个变量max_days,用于记录完成所有工作所需的最短天数。
4.遍历需求工作量列表,对于每个需求工作量Xm,将其分配给当前最空闲的员工(即工作时间最短的员工),并更新该员工的工作时间。
5.在每次分配完一个需求后,检查是否已经分配完所有需求,如果是,则更新max_days为当前最大员工工作时间;否则,继续分配下一个需求。
6.最后输出max_days作为结果。
JS算法源码 const rl = require("readline").createInterface({ input: process.stdin }); const readline = async () => (await rl[Symbol.asyncIterator]().next()).value; (async function () { const balls = (await readline()).split(" ").map(Number); const n = parseInt(await readline()); console.log(getResult(balls, n)); })(); function getResult(balls, n) { balls.sort((a, b) => b - a); let min = balls[0]; let max = balls.reduce((a, b) => a + b); let ans = max; while (min <= max) { const mid = (min + max) >> 1; if (check(balls, 0, new Array(n).fill(0), mid)) { ans = mid; max = mid - 1; } else { min = mid + 1; } } return ans; } function check(balls, index, buckets, limit) { if (index === balls.length) return true; const selected = balls[index]; for (let i = 0; i < buckets.length; i++) { if (i > 0 && buckets[i] === buckets[i - 1]) continue; if (selected + buckets[i] <= limit) { buckets[i] += selected; if (check(balls, index + 1, buckets, limit)) return true; buckets[i] -= selected; } } return false; }
Java算法源码 import java.util.Arrays; import java.util.Scanner; public class Main { static Integer[] balls; static int n; public static void main(String[] args) { Scanner sc = new Scanner(System.in); balls = Arrays.stream(sc.nextLine().split(" ")).map(Integer::parseInt).toArray(Integer[]::new); n = Integer.parseInt(sc.nextLine()); System.out.println(getResult()); } public static int getResult() { Arrays.sort(balls, (a, b) -> b - a); int min = balls[0]; int max = Arrays.stream(balls).reduce(Integer::sum).get(); int ans = max; while (min <= max) { int mid = (min + max) >> 1; if (check(0, new int[n], mid)) { ans = mid; max = mid - 1; } else { min = mid + 1; } } return ans; } public static boolean check(int index, int[] buckets, int limit) { if (index == balls.length) return true; int selected = balls[index]; for (int i = 0; i < buckets.length; i++) { if (i > 0 && buckets[i] == buckets[i - 1]) continue; if (selected + buckets[i] <= limit) { buckets[i] += selected; if (check(index + 1, buckets, limit)) return true; buckets[i] -= selected; } } return false; } }
Python算法源码 balls = list(map(int, input().split())) n = int(input()) def check(index, buckets, limit): if index == len(balls): return True selected = balls[index] for i in range(len(buckets)): if i > 0 and buckets[i] == buckets[i - 1]: continue if selected + buckets[i] <= limit: buckets[i] += selected if check(index + 1, buckets, limit): return True buckets[i] -= selected return False def getResult(): balls.sort(reverse=True) low = balls[0] high = sum(balls) ans = high while low <= high: mid = (low + high) >> 1 if check(0, [0] * n, mid): ans = mid high = mid - 1 else: low = mid + 1 return ans print(getResult())
C算法源码 #include <stdio.h> #include <stdlib.h> #define MAX_M 30 #define MAX_N 10 #define TRUE 1 #define FALSE 0 int balls[MAX_M]; int balls_size = 0; int buckets_size; int check(int index, int buckets[], int limit) { if (index == balls_size) return TRUE; int selected = balls[index]; for (int i = 0; i < buckets_size; i++) { if (i > 0 && buckets[i] == buckets[i - 1]) continue; if (selected + buckets[i] <= limit) { buckets[i] += selected; if (check(index + 1, buckets, limit)) { return TRUE; } buckets[i] -= selected; } } return FALSE; } int sum() { int sum = 0; for (int i = 0; i < balls_size; i++) sum += balls[i]; return sum; } int cmp(const void *a, const void *b) { return *((int *) b) - *((int *) a); } int main() { while (scanf("%d", &balls[balls_size++])) { if (getchar() != ' ') break; } scanf("%d", &buckets_size); qsort(balls, balls_size, sizeof(int), cmp); int min = balls[0]; int max = sum(); int ans = max; while (min <= max) { int mid = (min + max) >> 1; int buckets[MAX_N] = {0}; if (check(0, buckets, mid)) { ans = mid; max = mid - 1; } else { min = mid + 1; } } printf("%d\n", ans); return 0; }