Efficient Delivery

All submissions for this problem are available.
Read problems statements in Mandarin Chinese and Russian
Sergey works as a delivery boy. Every day he has a number of deliveries to be completed.
Sergey lives in Bytetown  a city, consisting of N junctions, and M roads connecting these junctions. More precisely, the i^{th} road connects the junctions with the numbers X_{i} and Y_{i}. Each of these roads has its own (positive) length Z_{i}, and no road connects a junction to itself.
There are K delivery orders for Sergey. Each order is denoted with four integers: A_{j}, B_{j}, V_{j} and C_{j}. It means that for the j^{th} order Sergey has to pick a parcel of weight V_{j} at the junction numbered A_{j} and deliver it to the junction numbered B_{j}. After the parcel is delivered, Sergey gets C_{j} burles of reward. Sergey can't complete the same order twice or more times.
So, the business day for Sergey looks as follows: he remembers the whole list of orders and starts at his initial junction numbered S. Then, he travel the streets of the city via his bike. If Sergey visits a junction, where there is one or more orders available he can either take some of the orders and load the corresponding parcels on his bike, or not take anything now but do it later, if he appears at the same junction again. When Sergey visits a junction which is the destination for some of the parcels that are currently with him, he can complete these orders and unload the corresponding parcels.
Sadly, Sergey is not infinitely strong. Though he can carry any number of parcels at once, if the total weight of the parcels exceeds W Sergey won't be able to carry them on his bike. Sergey also can't take a parcel and not deliver it because there will be complains, hurting Sergey's reputation.
The bike has enough fuel to travel only for D units of distance. It is not necessary for Sergey to get back to his initial junction because he has a lot of friends in Bytetown who will drop him back after the party (Sorry, you're not invited). It is clear that under the given constraints, Sergey might be not able to deliver all the parcels. So your task is to plan the delivery for Sergey in such a way that:
 He moves only by streets
 The total travelled distance doesn't exceed D.
 There are no undelivered parcels with Sergey by the end of the delivery. In other words, all the parcels that Sergey has picked up should be delivered.
 At any time, Sergey does not carry more than W units of weight.
 The most important: the profit Sergey gets by the end of the day is maximized. You don't have to find the optimal delivery, but the more profit there is, the more are the points you'll get.
Input
The first line contains two space separated integers N and M denoting the number of junctions and the number of streets.
Each of the following M lines contain three space separated integers X_{i}, Y_{i} and Z_{i}, denoting that the i^{th} street of length Z_{i} connects the X_{i}^{th} and the Y_{i}^{th} junctions. All the streets are bidirectional.
The next line contains a single integer K denoting the number of orders.
Each of the following K lines contain four spaceseparated integers A_{j}, B_{j}, V_{j} and C_{j}, denoting that for the j^{th} order, the parcel should be picked up at the A_{j}^{th} junction, left at the B_{j}^{th} junction, has the weight of V_{j}, and when delivered, gives C_{j} burles of profit.
The following line contains three spaceseparated integers S, D and W denoting the starting junction, the maximal distance that Sergey can travel and the maximum weight he can carry at any time, respectively.
Output
The first line of output should contain a single integer Op, denoting the number of operations. Each of the following Op lines should contain the operation code and the parameter. The following are the descriptions of the operation codes and the meanings of the parameters:
 0 P. Here 0 is the operation code and P is the parameter. The meaning is that Sergey should move from his current junction to the adjacent junction P. In case there is no direct onestreet connection between Sergey's current position and the junction P, you will receive a wrong answer verdict. In case Sergey is unable to reach the junction P because the amount of fuel is too low, you will also get wrong answer verdict.
 1 P. Here 1 is the operation code and P is the parameter. The meaning is that Sergey should takes the order numbered P in the orders' list. In case this order was already taken or Sergey is not in the junction which is the pickup junction for the order, you will get a wrong answer verdict. In case the weight that Sergey will have to carry exceeds W after taking this order, you'll also get a wrong answer verdict.
 2 P. Here 2 is the operation code and P is the parameter. The meaning is that Sergey should deliver the parcel for the order number P in 1based orders' list. In case Sergey doesn't carry the parcel from this order, you'll receive wrong answer verdict.
You can also get a wrong answer if your output doesn't satisfy the format above or some parcels picked up by Sergey are left undelivered. If all the requirements are satisfied, you'll receive an accepted verdict ;).
Test data and scoring
There are 20 test cases in total. They are split into 5 groups, 4 tests in each.
 In all the test cases:
 1 ≤ X_{i}, Y_{i}, A_{j}, B_{j}, S ≤ N
 1 ≤ V_{j} ≤ W
 1 ≤ C_{j} ≤ 10^{6}
 It is possible to reach any junction from the initial junction S via the streets.
 The streets are bidirectional.
 Each pair of junctions is connected with no more that one street.
 Group 1:
 1 ≤ N ≤ 100
 1 ≤ M, K ≤ 1000
 1 ≤ D, W ≤ 10000
 1 ≤ Z_{i} ≤ 100, for all the streets.
 Group 2:
 1 ≤ N ≤ 1000
 1 ≤ M, K ≤ 1000
 1 ≤ D ≤ 10000
 1 ≤ W ≤ 10^{6}
 Z_{i} = 1 for all the streets.
 Group 3:
 1 ≤ N ≤ 10000
 M = N1
 1 ≤ K ≤ 10000
 1 ≤ D ≤ 10^{5}
 1 ≤ W ≤ 10000
 1 ≤ Z_{i} ≤ 100, for all the streets
 Group 4:
 1 ≤ N ≤ 10^{5}
 1 ≤ M, K ≤ 10^{5}
 1 ≤ D ≤ 10^{5}
 1 ≤ W ≤ 10^{5}
 1 ≤ Z_{i} ≤ 10, for all the streets
 Group 5:
 1 ≤ N ≤ 10^{5}
 1 ≤ M, K ≤ 10^{5}
 1 ≤ D ≤ 10^{5}
 1 ≤ W ≤ 10^{9}
 1 ≤ Z_{i} ≤ 10, for all the streets
The score for a single test case equals the total profit of Sergey. The score for the whole task is the sum of the scores among all the test cases.
During the contest, you won't see the score of your solution on the complete test set. Instead of this, you will see the score, based on 4 test cases: one test case for each of groups 14. You will also see whether your program got accepted verdict on all the test cases (not only these 4), but time and memory stats will be reported based only on four premilinary test cases.
Example
Input: 5 5 1 2 1 2 3 2 1 4 1 4 5 1 5 2 3 3 1 4 5 10 2 5 6 15 4 1 10 10 1 5 12 Output: 7 1 1 0 2 1 2 0 5 2 2 0 4 2 1
Explanation
Sergey starts at the junction numbered the 1^{st}. Before moving on, he takes the order numbered the 1^{st}. Then he moves to the 2^{nd} junction and there he takes the 2^{nd} order. Then, Sergey moves to the 5^{th} junction, where he delivers the parcel from the 2^{nd} order. Finally, he moves to the 4^{th} junction, where he delivers the parcel from the 1^{th} order.
The total profit that Sergey receives from this delivery is (10 + 15) = 25 burles. Hence, this answer will get the score of 25.
Author:  xcwgf666 
Editorial  http://discuss.codechef.com/problems/EFFDELIV 
Tags  challenge, dijkstra, knapsack, oct15, xcwgf666 
Date Added:  12092015 
Time Limit:  5 sec 
Source Limit:  50000 Bytes 
Languages:  C, CPP14, JAVA, PYTH, PYTH 3.5, PYPY, CS2, PAS fpc, PAS gpc, RUBY, PHP, GO, NODEJS, HASK, SCALA, D, PERL, FORT, WSPC, ADA, CAML, ICK, BF, ASM, CLPS, PRLG, ICON, SCM qobi, PIKE, ST, NICE, LUA, BASH, NEM, LISP sbcl, LISP clisp, SCM guile, JS, ERL, TCL, PERL6, TEXT, SCM chicken, CLOJ, FS 
Comments
 Please login at the top to post a comment.
SUCCESSFUL SUBMISSIONS
Fetching successful submissions
HELP
If you are still having problems, see a sample solution here. 