车间调度问题

问题描述 六工四机问题

一个车间有四台机器，要求加工6个作业

分工序，每个机器加工不同的工序

//深度搜索算法
#include <stdio.h>
#include <string.h>
#include <algorithm>
using namespace std;
int totalStep, minTime;
int n, m;
int Step[105];
struct Job
{
    int machine;
    int len;
} job[105][105];
int jobEnd[105][105];
int jobStep[105];
int machineWorkTime[105];
struct Recording
{
    int start;
    int ed;
    int job;
    int machine;
} best[105], now[105];
void init()
{
    totalStep = 0;
    minTime = 999999999;
    memset(jobStep, 0, sizeof(jobStep));
    memset(machineWorkTime, 0, sizeof(machineWorkTime));
    memset(best, 0, sizeof(best));
    memset(jobEnd, -1, sizeof(jobEnd));
}
void dfs(int step, int time)
{
    if (time >= minTime)
        return;
    if (step == totalStep)
    {
        minTime = time;
        for (int i = 0; i < totalStep; i++)
            best[i] = now[i];
        /*int gantt[105][105];
        for (int i=0;i<totalStep;i++)
        {
            for (int j=best[i].start;j<best[i].ed;j++)
            {
                gantt[best[i].machine][j]=best[i].job;
            }
        }
        for (int i=0;i<m;i++)
            for (int j=0;j<minTime;j++)
                printf("%d%c",gantt[i][j],j==minTime-1?'\n':' ');
            printf("\n");*/
        return;
    }

    for (int i = 0; i < n; i++)
    {
        int j = jobStep[i];
        if (j >= Step[i])
            continue;
        int thisMachine = job[i][j].machine;
        now[step].machine = thisMachine;
        now[step].job = i + 1;
        int temp = machineWorkTime[thisMachine];
        int beginTime = max(jobEnd[i][j - 1], machineWorkTime[thisMachine]);
        now[step].start = beginTime;
        machineWorkTime[thisMachine] = beginTime + job[i][j].len;
        jobEnd[i][j] = now[step].ed = machineWorkTime[thisMachine];
        jobStep[i]++;
        // printf("%d %d %d\n",step,i,j);
        dfs(step + 1, max(time, machineWorkTime[thisMachine]));
        jobStep[i]--;
        machineWorkTime[thisMachine] = temp;
    }
}
int main()
{
    printf("输入工作数量：");
    scanf("%d", &n);
    printf("输入机器数量：");
    scanf("%d", &m);
    init();
    for (int i = 0; i < n; i++)
    {
        printf("输入工作%d的步骤数量：", i);
        scanf("%d", &Step[i]);
        totalStep += Step[i];
        for (int j = 0; j < Step[i]; j++)
        {
            printf("步骤%d运行于哪个机器，消耗时间：", j + 1);
            scanf("%d%d", &job[i][j].machine, &job[i][j].len);
        }
    }
    dfs(0, 0);
    printf("最短时间为：%d\n", minTime);

    int gantt[105][105];
    for (int i = 0; i < totalStep; i++)
    {
        for (int j = best[i].start; j < best[i].ed; j++)
        {
            gantt[best[i].machine][j] = best[i].job;
        }
    }

    for (int i = 0; i < m; i++)
        for (int j = 0; j < minTime; j++)
            printf("%d%c", gantt[i][j], j == minTime - 1 ? '\n' : ' ');
}

标签：车间,int,调度,问题,++,job,machineWorkTime,best,105
来源： https://www.cnblogs.com/W-xzg/p/16161117.html