算法导论15章装配线实现(c++)

#include "stdafx.h"
#include <iostream>
#include <math.h>

using namespace std;

#define N 6


int f1[N] = {0},f2[N] = {0},f_final = 0;//f1数组记录装配线1中某装配站的最小时间，f_final记录完工总时间
int l1[N-1] = {0},l2[N-1] = {0},l_final = 0;//l1数组记录装配线1中某装配站输入从1还是2中来，故值取1或者2，l_final记录最后从x1还是x2完成，每条线第一个站不记录
void Fastest_way (int a[2][N],int e1,int e2,int x1,int x2,int t1[],int t2[])//查找最快装配时间，用f1、f2和l1、12记录
{
    f1[0] = e1 + a[0][0];
    f2[0] = e2 + a[1][0];
    for (int j = 1; j < N; j++)
    {
        if ( f1[j-1] + a[0][j] <= f2[j-1] + t2[j-1] + a[0][j])
        {
            f1[j] = f1[j-1] + a[0][j];
            l1[j-1] = 1;
        }
        else
        {
            f1[j] = f2[j-1] + t2[j-1] + a[0][j];
            l1[j-1] = 2;
        }

        if ( f2[j-1] + a[1][j] <= f1[j-1] + t1[j-1] + a[1][j])
        {
            f2[j] = f2[j-1] + a[1][j];
            l2[j-1] = 2;
        }
        else
        {
            f2[j] = f1[j-1] + t1[j-1] + a[1][j];
            l2[j-1] = 1;
        }
        
    }

    if (f1[N-1] + x1 <= f2[N-1] + x2)
    {
        f_final = f1[N-1] + x1;
        l_final = 1;
    }
    else
    {
        f_final = f2[N-1] + x2;
        l_final = 2;
    }
}

void Print_stations()//逆序查找输出所经过的装配站
{
    int i = l_final;
    cout<<"line "<<i<<",station "<<N<<endl;
    for (int j = N-1; j > 0; j--)
    {
        if (i==1)
            i = l1[j-1];
        else
            i = l2[j-1];
        cout<<"line "<<i<<",station "<<j<<endl;
    }
}

int main()
{
    int e1 = 2, e2 = 4;
    int x1 = 3, x2 = 2;
    int t1[N-1] = {2, 3, 1, 3, 4};
    int t2[N-1] = {2, 1, 2, 2, 1};
    int a[2][N] = {{7, 9, 3, 4, 8, 4},{8, 5, 6, 4, 5, 7}};
    Fastest_way (a, e1, e2, x1, x2, t1, t2);
    cout<<f_final<<endl;
    Print_stations();
    system("pause");
    return 0;
}