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某公司研发了一款高性能AI处理器。每台物理设备具备8颗AI处理器，编号分别为0、1、2、3、4、5、6、7。

编号0-3的处理器处于同一个链路中，编号4-7的处理器处于另外一个链路中，不同链路中的处理器不能通信。

现给定服务器可用的处理器编号数组array，以及任务申请的处理器数量num，找出符合下列亲和性调度原则的芯片组合。

如果不存在符合要求的组合，则返回空列表。

亲和性调度原则：

-如果申请处理器个数为1，则选择同一链路，剩余可用的处理器数量为1个的最佳，其次是剩余3个的为次佳，然后是剩余2个，最后是剩余4个。

-如果申请处理器个数为2，则选择同一链路剩余可用的处理器数量2个的为最佳，其次是剩余4个，最后是剩余3个。

-如果申请处理器个数为4，则必须选择同一链路剩余可用的处理器数量为4个。

-如果申请处理器个数为8，则申请节点所有8个处理器。

提示：

任务申请的处理器数量只能是1、2、4、8。
编号0-3的处理器处于一个链路，编号4-7的处理器处于另外一个链路。
处理器编号唯一，且不存在相同编号处理器。
输入描述

输入包含可用的处理器编号数组array，以及任务申请的处理器数量num两个部分。

第一行为array，第二行为num。例如：

[0, 1, 4, 5, 6, 7]

表示当前编号为0、1、4、5、6、7的处理器可用。任务申请1个处理器。

0 <= array.length <= 8

0 <= array[i] <= 7

num in [1, 2, 4, 8]

输出描述

输出为组合列表，当array=[0，1，4，5，6，7]，num=1 时，输出为[[0], [1]]。

示例1 输入输出示例仅供调试，后台判题数据一般不包含示例

输入

[0, 1, 4, 5, 6, 7]

输出

[[0], [1]]

说明

根据第一条亲和性调度原则，在剩余两个处理器的链路（0, 1, 2, 3）中选择处理器。

由于只有0和1可用，则返回任意一颗处理器即可。

Java 代码

import java.util.Scanner;
import java.util.*;
import java.util.stream.Stream;
import java.util.stream.Collectors;class Main {public static void main(String[] args) {// 处理输入Scanner in = new Scanner(System.in);Integer[] cores = Arrays.stream(in.nextLine().split("[\\[\\]\\,\\s]")).filter(str -> !"".equals(str)).map(Integer::parseInt).toArray(Integer[]::new);int target = in.nextInt();//初始化两个链路剩余可用的处理器ArrayList<Integer> processors_1 = new ArrayList<>();ArrayList<Integer> processors_2 = new ArrayList<>();Arrays.sort(cores, (a, b) -> a - b);for (Integer core : cores) {if (core < 4) {processors_1.add(core);} else {processors_2.add(core);}}ArrayList<ArrayList<Integer>> result = new ArrayList<>();int length_1 = processors_1.size();int length_2 = processors_2.size();switch (target) {case 1:if (length_1 == 1 || length_2 == 1) {if (length_1 == 1) dfs(processors_1, 0, 1, new ArrayList<>(), result);if (length_2 == 1) dfs(processors_2, 0, 1, new ArrayList<>(), result);} else if (length_1 == 3 || length_2 == 3) {if (length_1 == 3) dfs(processors_1, 0, 1, new ArrayList<>(), result);if (length_2 == 3) dfs(processors_2, 0, 1, new ArrayList<>(), result);} else if (length_1 == 2 || length_2 == 2) {if (length_1 == 2) dfs(processors_1, 0, 1, new ArrayList<>(), result);if (length_2 == 2) dfs(processors_2, 0, 1, new ArrayList<>(), result);} else if (length_1 == 4 || length_2 == 4) {if (length_1 == 4) dfs(processors_1, 0, 1, new ArrayList<>(), result);if (length_2 == 4) dfs(processors_2, 0, 1, new ArrayList<>(), result);}break;case 2:if (length_1 == 2 || length_2 == 2) {if (length_1 == 2) dfs(processors_1, 0, 2, new ArrayList<>(), result);if (length_2 == 2) dfs(processors_2, 0, 2, new ArrayList<>(), result);} else if (length_1 == 4 || length_2 == 4) {if (length_1 == 4) dfs(processors_1, 0, 2, new ArrayList<>(), result);if (length_2 == 4) dfs(processors_2, 0, 2, new ArrayList<>(), result);} else if (length_1 == 3 || length_2 == 3) {if (length_1 == 3) dfs(processors_1, 0, 2, new ArrayList<>(), result);if (length_2 == 3) dfs(processors_2, 0, 2, new ArrayList<>(), result);}break;case 4:if (length_1 == 4 || length_2 == 4) {if (length_1 == 4) result.add(processors_1);if (length_2 == 4) result.add(processors_2);}break;case 8:if (length_1 == 4 && length_2 == 4) {result.add(Stream.concat(processors_1.stream(), processors_2.stream()).collect(Collectors.toCollection(ArrayList<Integer>::new)));}break;}System.out.println(result.toString());}public static void dfs(ArrayList<Integer> cores,int index,int level,ArrayList<Integer> path,ArrayList<ArrayList<Integer>> res) {if (path.size() == level) {res.add((ArrayList<Integer>) path.clone());return;}for (int i = index; i < cores.size(); i++) {path.add(cores.get(i));// 逐个往后找合适的组合dfs(cores, i + 1, level, path, res);path.remove(path.size() - 1);}}
}

Python代码

import functools
import collections
import math
from itertools import combinations
from re import match
import copyclass TreeNode:def __init__(self, val=0, left=None, right=None):self.val = valself.left = leftself.right = right#并查集模板
class UF:def __init__(self, n=0):self.count = nself.item = [0 for x in range(n+1)]for i in range(n):self.item[i] = idef find(self, x):if (x != self.item[x]):self.item[x] = self.find(self.item[x])return 0return xdef union_connect(self, x, y):x_item = self.find(x)y_item = self.find(y)if (x_item != y_item):self.item[y_item] = x_itemself.count-=1# 处理输入
v = [int(x) for x in input()[1:-1].split(",")]#初始化两个链路剩余可用的处理器
processors_1 = []
processors_2 = []
for x in v:if (x >= 4):processors_2.append(x)else:processors_1.append(x)length_1 = len(processors_1)
length_2 = len(processors_2)# 申请的处理器个数
apply_num = int(input())#满足条件结果组合
result = []def dfs(batch_processor,index, level, path):if (len(path) == level):result.append(copy.copy(path))return for i in range(index, len(batch_processor)):path.append(batch_processor[i])# 逐个往后找合适的组合dfs(batch_processor, i + 1, level, path)path.pop()def get_combo(batch_processor, num):path=[]dfs(batch_processor, 0, num, path)if (apply_num == 1):# 原则1if (length_1 == 1 or length_2 == 1):if (length_1 == 1): get_combo(processors_1, 1)if (length_2 == 1): get_combo(processors_2, 1)elif (length_1 == 3 or length_2 == 3):if (length_1 == 3): get_combo(processors_1, 1)if (length_2 == 3): get_combo(processors_2, 1)elif (length_1 == 2 or length_2 == 2):if (length_1 == 2): get_combo(processors_1, 1)if (length_2 == 2): get_combo(processors_2, 1)elif (length_1 == 4 or length_2 == 4):if (length_1 == 4): get_combo(processors_1, 1)if (length_2 == 4): get_combo(processors_2, 1)elif (apply_num == 2):# 原则2if (length_1 == 2 or length_2 == 2):if (length_1 == 2): get_combo(processors_1, 2)if (length_2 == 2): get_combo(processors_2, 2)elif (length_1 == 4 or length_2 == 4):if (length_1 == 4): get_combo(processors_1, 2)if (length_2 == 4): get_combo(processors_2, 2)elif (length_1 == 3 or length_2 == 3):if (length_1 == 3): get_combo(processors_1, 2)if (length_2 == 3): get_combo(processors_2, 2)elif (apply_num == 4):# 原则3if (length_1 == 4 or length_2 == 4):if (length_1 == 4): result.append(processors_1)if (length_2 == 4): result.append(processors_2)elif (apply_num == 8):# 原则4if (length_1 == 4 and length_2 == 4):processors_2 = processors_2 + processors_1result.append(processors_2)result = [str(x) for x in result]
print("[" +", ".join(result)+ "]")

JS代码

let result = []function dfs(batch_processor,index, level, path){if (path.length == level){result.push([...path])return }for (let i=index;i<batch_processor.length;i++){path.push(batch_processor[i])// 逐个往后找合适的组合dfs(batch_processor, i + 1, level, path)path.pop()}
}function get_combo(batch_processor, num){let path=[]dfs(batch_processor, 0, num, path)
}function main(v, apply_num) {//初始化两个链路剩余可用的处理器let processors_1 = []let processors_2 = []for (let x of v){if (x >= 4)processors_2.push(x)elseprocessors_1.push(x)}let length_1 = processors_1.lengthlet length_2 = processors_2.lengthif (apply_num == 1){// 原则1if (length_1 == 1 || length_2 == 1){if (length_1 == 1)get_combo(processors_1, 1)if (length_2 == 1)get_combo(processors_2, 1)}else if (length_1 == 3 || length_2 == 3){if (length_1 == 3)get_combo(processors_1, 1)if (length_2 == 3)get_combo(processors_2, 1)}else if (length_1 == 2 || length_2 == 2){if (length_1 == 2)get_combo(processors_1, 1)if (length_2 == 2)get_combo(processors_2, 1)}else if (length_1 == 4 || length_2 == 4){if (length_1 == 4)get_combo(processors_1, 1)if (length_2 == 4)get_combo(processors_2, 1)}}else if (apply_num == 2){// 原则2if (length_1 == 2 || length_2 == 2){if (length_1 == 2)get_combo(processors_1, 2)if (length_2 == 2)get_combo(processors_2, 2)}else if (length_1 == 4 || length_2 == 4){if (length_1 == 4)get_combo(processors_1, 2)if (length_2 == 4)get_combo(processors_2, 2)}else if (length_1 == 3 || length_2 == 3){if (length_1 == 3)get_combo(processors_1, 2)if (length_2 == 3)get_combo(processors_2, 2)}}else if (apply_num == 4){// 原则3if (length_1 == 4 || length_2 == 4){if (length_1 == 4)result.push(processors_1)if (length_2 == 4)result.push(processors_2)}}   else if (apply_num == 8){// 原则4if (length_1 == 4 && length_2 == 4){processors_2 = processors_2 + processors_1result.push(processors_2)}}console.log(result)}main([0, 1, 4, 5, 6, 7], 1)