Dynamic Programming: assembly line scheduling problem (algorithm guide ...

assembly line Scheduling problem
Problem Description:
Colonel Motor Company produces cars in factories with two assembly lines, and a car site enters every assembly line from the start,
Install the parts in some assembly stations, assuming that each line has n assembly stations and is numbered,..., N. Start entering two lines to consume
E1, E2 time, respectively, represents the time to enter two assembly lines, each assembly station needs to consume a certain amount of time, and in each
Jump between assembly stations takes a certain amount of time timei[j], your task is to calculate the shortest time to complete the assembly. (Because each line
The same numbered assembly stations are assembled with the same parts, but the time varies depending on the technology.)

Input:
N: Number of assembly stations, E1,e2: The time to enter the assembly line, X1,X2: The time of the assembly line,
The time of the N assembly station of the first line, the time of the second line n the Assembly station, the time of the n-1 of the two lines
Output:
The shortest time to complete all assembly.

Sample input:
6
2 4 3 2
7 9 3 4 8 4
8 5 6 4 5 7
2 3 1) 3 4
2 1 2) 2 1

Sample output:
38

Problem Solving Ideas:
1. The classic problem has a new harvest every time it is analyzed.
2. Now we assume the state is: Dp[i][j]: Indicates the minimum time for the J Assembly station on line I.
3. Dynamic equation: dp[i][j] = min (dp[i][j-1] + a[i][j], dp[i^1][j]+time[i^1][j-1]+a[i][j]);
(i^1 means take another line)

Code:
#include <cstdio>
#include <iostream>
#include <cstring>
using namespace Std;
#define MAX 100

int n;
int F1[max], F2[max];
int A1[max], A2[max];
int L1[max], L2[max];
int e1, e2, X1, X2;
int Time1[max], Time2[max];

int DP ()
{
int result;
F1[1] = e1 + a1[1];
F2[1] = e2 + a2[1];
for (int i = 2; I <= n; ++i)
{
if (F1[i-1] + a1[i] <= f2[i-1] + time2[i-1] + a1[i])
{
F1[i] = F1[i-1]+a1[i];
L1[i] = 1;
}
Else
{
F1[i] = F2[i-1]+time2[i-1]+a1[i];
L1[i] = 2;
}

if (F2[i-1] + a2[i] <= f1[i-1] + time1[i-1] + a2[i])
{
F2[i] = F2[i-1]+a2[i];
L2[i] = 1;
}
Else
{
F2[i] = F1[i-1]+time1[i-1]+a2[i];
L2[i] = 2;