SHPATH - The Shortest Path

You are given a list of cities. Each direct connection between two cities has its transportation cost (an integer bigger than 0). The goal is to find the paths of minimum cost between pairs of cities. Assume that the cost of each path (which is the sum of costs of all direct connections belonging to this path) is at most 200000. The name of a city is a string containing characters a, ..., z and is at most 10 characters long.

Input

s [the number of tests <= 10]
n [the number of cities <= 10000]
NAME [city name]
p [the number of neighbours of city NAME]
nr cost [nr - index of a city connected to NAME (the index of the first city is 1)]
        [cost - the transportation cost]
r [the number of paths to find <= 100]
NAME1 NAME2 [NAME1 - source, NAME2 - destination]
[empty line separating the tests]

Output

cost [the minimum transportation cost from city NAME1 to city NAME2 (one per line)]

Example

Input:
1
4
gdansk
2
2 1
3 3
bydgoszcz
3
1 1
3 1
4 4
torun
3
1 3
2 1
4 1
warszawa
2
2 4
3 1
2
gdansk warszawa
bydgoszcz warszawa

Output:
3
2

Warning: large Input/Output data, be careful with certain languages


Added by:Darek Dereniowski
Date:2004-05-10
Time limit:1.5s
Source limit:50000B
Memory limit:1536MB
Cluster: Cube (Intel G860)
Languages:All
Resource:DASM Programming League 2003 (problemset 11)

hide comments
2019-04-19 22:14:44 Narendra yadav
@alphacoder101:
How can you .... "Stop Dijkstra if the destination is arrived i.e if current node is destination". Won't this give WA (non-optimal) ?

Last edit: 2019-04-19 22:15:00
2019-04-07 08:59:40
Phew!! got AC with 1.29 sec..
2019-02-06 06:29:38
AC in one go.
Consider using :
1. 1 - indexed list
2. Min - Priority queue ( Priority Queue with negative weights / Multiset )
3. Stop Dijkstra if the destination is arrived i.e if current node is destination
4. Use maps to convert City names to indexes
2019-01-03 18:25:03
Simpe Djikstra,using priority_queue in c++.
2018-12-25 20:24:35
That was close! AC in 0.94
2018-12-16 11:45:16
Don't use set , use priority queue to save yourself from TLE
2018-10-08 00:15:39
TIP: Use the priority_queue implementation of Dijkstra to avoid TLE instead of set
2018-09-24 08:05:51
Hello World in Dijkstra
2018-06-06 08:17:48
Don' t use Java. You will get tle in this question no matter what optimization you do.
2018-05-28 20:23:32
very imp note: make sure to clear your graph after each test case, will save you a wa :)
© Spoj.com. All Rights Reserved. Spoj uses Sphere Engine™ © by Sphere Research Labs.