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### 10740 - Not the Best

Posted: Tue Oct 19, 2004 8:41 am
WA ing

Thanks !

### ^^

Posted: Tue Oct 19, 2004 10:23 am
I got WA, too.

My solution is BFS with Priority Queue. Is Right?

Posted: Tue Oct 19, 2004 1:11 pm
My input:

Code: Select all

``````5 6
1 5 2
1 2 1
2 5 10
1 3 1
3 5 10
1 4 1
4 5 10

5 6
1 5 3
1 2 1
2 5 10
1 3 1
3 5 10
1 4 1
4 5 10

5 6
1 5 4
1 2 1
2 5 10
1 3 1
3 5 10
1 4 1
4 5 10

5 6
1 5 5
1 2 1
2 5 10
1 3 1
3 5 10
1 4 1
4 5 10
3 3

1 3 4

1 3 3

1 2 4

2 3 5

5 6

5 2 5

1 2 2

2 5 4

3 2 3

4 3 1

5 1 3

5 4 2

2 2

1 2 3

1 2 5

2 2 2

5 6

5 2 2

1 2 2

2 5 4

3 2 3

4 3 1

5 1 3

5 4 2
5 6

5 2 3

1 2 2

2 5 4

3 2 3

4 3 1

5 1 3

5 4 2
5 6

5 2 4

1 2 2

2 5 4

3 2 3

4 3 1

5 1 3

5 4 2
5 6

5 2 5

1 2 2

2 5 4

3 2 3

4 3 1

5 1 3

5 4 2
5 6

5 2 6

1 2 2

2 5 4

3 2 3

4 3 1

5 1 3

5 4 2
5 6

5 2 7

1 2 2

2 5 4

3 2 3

4 3 1

5 1 3

5 4 2
5 6

5 2 8

1 2 2

2 5 4

3 2 3

4 3 1

5 1 3

5 4 2
5 6

5 2 9

1 2 2

2 5 4

3 2 3

4 3 1

5 1 3

5 4 2
5 6

5 2 10

1 2 2

2 5 4

3 2 3

4 3 1

5 1 3

5 4 2

0 0

``````
My output:

Code: Select all

``````11
11
-1
-1
-1
15
9
6
14
15
15
16
23
24
24
24``````
PLZ point out my errors,

Thanks !

Posted: Tue Oct 19, 2004 3:05 pm
My AC program gives same output.
Have you considered more than one edge from between 2 vertices?
I am not sure if the judge input has such case, but the problem statment doesn't avoid that.

Posted: Tue Oct 19, 2004 3:32 pm
how about the case when the weight of the edge is 0. the problem says the weights are >=0;

Posted: Tue Oct 19, 2004 3:41 pm
.. wrote:My AC program gives same output.
Have you considered more than one edge from between 2 vertices?
I am not sure if the judge input has such case, but the problem statment doesn't avoid that.
The judge input hasn't such case , becase I had checked it

Posted: Tue Oct 19, 2004 3:53 pm
abishek wrote:how about the case when the weight of the edge is 0. the problem says the weights are >=0;
what's the speciality of this case?

Posted: Tue Oct 19, 2004 8:27 pm
there is nothing special about that case. I was just wondering if you had made that assumption some where else in the code.

Posted: Wed Oct 20, 2004 7:31 am
I changed my algorithm to BFS and got AC now

Thanks !

### Help me!

Posted: Wed Oct 20, 2004 10:04 pm
I'm using a very naive algorithm: a simple BFS + priority queue (without checking if a node has been visited, always inserting in the queue). However, my algorithm doesn't seem to work. I was getting TLE, but now I'm getting WA. I realized that there can be zero-weighted cycles, which can cause my algo to infinite loop.

I could use some tips, inputs, or some reference to a classic kth shortest path algorithm.

### ^^

Posted: Thu Oct 21, 2004 5:03 am
TLE...Test Case
4 4
1 4 10
1 2 1
2 3 1
3 1 1
1 4 10000
BFS with Priority Queue can fall in local loop.

### Some ideas

Posted: Thu Oct 21, 2004 12:20 pm
Hello sozu,

Thank you for the test case. As I said in my previous post, a zero-sized loop could be even worse
4 4
1 4 10
1 2 0
2 3 0
3 1 0
1 4 10000
I tried some ideas like only inserting a node K times in the queue, but that didn't work. Could someone give some references to Kth shortest paths algorithms? I couldn't find anything really useful on the web. I would also appreciate some ideas on how to solve this problem (it's not like I want someones code, but just some ideas that could lead me to the right direction )

Posted: Thu Oct 21, 2004 12:45 pm
I did a simple and not too fast exhaustive search (dfs) keeping a top-k of distances from the source per node. The break-off condition for the search is when a distance outside the top-k is reached. At the end of the search the answer is in k-th place in the top-k of the target node.
A resonable speed-up (but not spectacular) was made by searching the nodes closest to the current node first (sorting the adjacency list). You can, off course, visit a node more than once during dfs.

Posted: Sat Oct 23, 2004 3:20 pm
Isn't there a reference for the kth shortest path after all?

Posted: Sat Oct 23, 2004 10:34 pm
What reference do you need? Isn't slightly modified Dijkstra (with counting the k-th shortest path to all the verticles) good enough?