华为OD机考算法题：高效的任务规划

题目部分

解读与分析

题目解读：

每台机器只有配置了才能执行。而在同一个时间段只能执行一台机器的配置（配置串行执行），在配置完成后，任务即可执行。
求出执行完所有任务的时间。

分析与思路：

对于每一组数据，可以采取贪心算法，遍历所有的组合情况，求出每种情况所需要的时间，经比较，输出时间最小的数字。

此算法的时间复杂度为 O()，空间复杂度为  O()。

代码实现

Java代码

import java.util.Scanner;
 
/**
 * 高效的任务规划
 * 
 * @since 2023.10.25
 * @version 0.1
 * @author Frank
 *
 */
public class EfficientTaskPlan {
    public static void main(String[] args) {
	Scanner sc = new Scanner(System.in);
	while (sc.hasNext()) {
	    String input = sc.nextLine();
	    int groupCnt = Integer.parseInt(input);
	    int[] outputValues = new int[ groupCnt ];
	    for( int i = 0; i < groupCnt; i ++ )
	    {
		input = sc.nextLine();
		int machineCnt = Integer.parseInt( input );
		int [][] taskInfo = new int[machineCnt][2];
		for( int j = 0; j < machineCnt; j ++ )
		{
		    input = sc.nextLine();
		    String[] eachMachineStr = input.split( " " );
		    int[] eachMachine = new int[2];
		    eachMachine[0] = Integer.parseInt( eachMachineStr[0] );
		    eachMachine[1] = Integer.parseInt( eachMachineStr[1] );
		    taskInfo[j] = eachMachine;
		}
		int[] usedFlag = new int[taskInfo.length];
		for( int j = 0; j < usedFlag.length; j ++ )
		{
		    usedFlag[j] = 0;
		}
		outputValues[i] = caculateEachGroupTaskPlan( usedFlag, taskInfo, 0 );
	    }
 
	    for( int i = 0; i < groupCnt; i ++ )
	    {
		System.out.println( outputValues[i] );
	    }
	}
    }
 
    private static int caculateEachGroupTaskPlan( int[] usedFlag, int [][] taskInfo, int curTask ) {
	int minTimeTake = Integer.MAX_VALUE;
	for( int i = 0; i < taskInfo.length; i ++ )
	{
	    if( usedFlag[i] == 1 )
	    {
		continue;
	    }
	    
	    int tmpConfig = taskInfo[i][0];
	    int tmpTask = taskInfo[i][1];
	    
	    usedFlag[i] = 1;
	    int curTimeTake = tmpConfig + caculateEachGroupTaskPlan( usedFlag, taskInfo, tmpTask );
	    usedFlag[i] = 0;
	    if( curTimeTake <= curTask )
	    {
		return curTask;
	    }
	    if( curTimeTake < minTimeTake )
	    {
		minTimeTake = curTimeTake;
	    }
	    
	}	
	if( minTimeTake < curTask || minTimeTake == Integer.MAX_VALUE )
	{
	    minTimeTake = curTask;
	}
	return minTimeTake;
    }
}

JavaScript代码

const rl = require("readline").createInterface({ input: process.stdin });
var iter = rl[Symbol.asyncIterator]();
const readline = async () => (await iter.next()).value;
void async function() {
    while (line = await readline()) {
        var groupCnt = parseInt(line);
        var outputValues = new Array();
        for (var i = 0; i < groupCnt; i++) {
            line = await readline();
            var machineCnt = parseInt(line);
            var taskInfo = new Array();
            for (var j = 0; j < machineCnt; j++) {
                line = await readline();
                var eachMachineStr = line.split(" ");
                var eachMachine = new Array();
                eachMachine[0] = parseInt(eachMachineStr[0]);
                eachMachine[1] = parseInt(eachMachineStr[1]);
                taskInfo[j] = eachMachine;
            }
            var usedFlag = new Array();
            for (var j = 0; j < groupCnt; j++) {
                usedFlag[j] = 0;
            }
            outputValues[i] = caculateEachGroupTaskPlan(usedFlag, taskInfo, 0);
        }
 
        for (var i = 0; i < groupCnt; i++) {
            console.log(outputValues[i]);
        }
 
    }
}();
 
function caculateEachGroupTaskPlan(usedFlag, taskInfo, curTask) {
    var minTimeTake = Number.MAX_VALUE;
    for (var i = 0; i < taskInfo.length; i++) {
        if (usedFlag[i] == 1) {
            continue;
        }
 
        var tmpConfig = taskInfo[i][0];
        var tmpTask = taskInfo[i][1];
 
        usedFlag[i] = 1;
        var curTimeTake = tmpConfig + caculateEachGroupTaskPlan(usedFlag, taskInfo, tmpTask);
        usedFlag[i] = 0;
        if (curTimeTake <= curTask) {
            return curTask;
        }
        if (curTimeTake < minTimeTake) {
            minTimeTake = curTimeTake;
        }
 
    }
    if (minTimeTake < curTask || minTimeTake == Number.MAX_VALUE) {
        minTimeTake = curTask;
    }
    return minTimeTake;
}

(完)