gs gs122 Dynamic Programming - Assembly Line Scheduling - Quiz No.1

 time_to_reach[2][3] = {{17, 2, 7}, {19, 4, 9}} time_spent[2][4] = {{6, 5, 15, 7}, {5, 10, 11, 4}} entry_time[2] = {8, 10} exit_time[2] = {10, 7} num_of_stations = 4

 time_to_reach[2][3] = {{17, 2, 7}, {19, 4, 9}} time_spent[2][4] = {{6, 5, 15, 7}, {5, 10, 11, 4}} entry_time[2] = {8, 10} exit_time[2] = {10, 7} num_of_stations = 4

 time_to_reach[2][3] = {{17, 2, 7}, {19, 4, 9}} time_spent[2][4] = {{6, 5, 15, 7}, {5, 10, 11, 4}} entry_time[2] = {8, 10} exit_time[2] = {10, 7} num_of_stations = 4

 #include<stdio.h> int get_min(int a, int b) { if(a<b) return a; return b; } int minimum_time_required(int reach[][3],int spent[][4], int *entry, int *exit, int n) { int t1[n], t2[n],i; t1[0] = entry[0] + spent[0][0]; t2[0] = entry[1] + spent[1][0]; for(i = 1; i < n; i++) { t1[i] = get_min(t1[i-1]+spent[0][i], t2[i-1]+reach[1][i-1]+spent[0][i]); __________; } return get_min(t1[n-1]+exit[0], t2[n-1]+exit[1]); }

 #include<stdio.h> int get_min(int a, int b) { if(a<b) return a; return b; } int minimum_time_required(int reach[][3],int spent[][4], int *entry, int *exit, int n) { int t1[n], t2[n], i; t1[0] = entry[0] + spent[0][0]; t2[0] = entry[1] + spent[1][0]; for(i = 1; i < n; i++) { t1[i] = get_min(t1[i-1]+spent[0][i], t2[i-1]+reach[1][i-1]+spent[0][i]); t2[i] = get_min(t2[i-1]+spent[1][i], t1[i-1]+reach[0][i-1]+spent[1][i]); } return get_min(t1[n-1]+exit[0], t2[n-1]+exit[1]); } int main() { int time_to_reach[][3] = {{6, 1, 5}, {2, 4, 7}}; int time_spent[][4] = {{6, 5, 4, 7}, {5, 10, 2, 6}}; int entry_time[2] = {5, 6}; int exit_time[2] = {8, 9}; int num_of_stations = 4; int ans = minimum_time_required(time_to_reach, time_spent, entry_time, exit_time, num_of_stations); printf("%d",ans); return 0; }