Help on a USACO problem

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Sanny
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Help on a USACO problem

Post by Sanny »

It is a problem of last month's USACO gold division. The description is:

Code: Select all


FJ's cows really hate getting wet so much that the mere thought of getting caught in the rain makes them shake in their hooves. They have decided to put a rain siren on the farm to let them know when rain is approaching. They intend to create a rain evacuation plan so that all the cows can get to shelter before the rain begins. Weather forecasting is not always correct, though. In order to minimize false alarms, they want to sound the siren as late as possible while still giving enough time for all the cows to get to some shelter. 

The farm has F (1 <= F <= 200) fields on which the cows graze. A set of P (1 <= P <= 1500) paths connects them. The paths are wide, so that any number of cows can traverse a path in either direction. 

Some of the farm's fields have rain shelters under which the cows can shield themselves. These shelters are of limited size, so a single shelter might not be able to hold all the cows. Fields are small compared to the paths and require no time for cows to traverse. 

Compute the minimum amount of time before rain starts that the siren must be sounded so that every cow can get to some shelter.

Input 

* Line 1: Two space-separated integers: F and P 

* Lines 2..F+1: Two space-separated integers that describe a field. The first integer (range: 0..1000) is the number of cows in that field. The second integer (range: 0..1000) is the number of cows the shelter in that field can hold. Line i+1 describes field i. 

* Lines F+2..F+P+1: Three space-separated integers that describe a path. The first and second integers (both range 1..F) tell the fields connected by the path. The third integer (range: 1..1,000,000,000) is how long any cow takes to traverse it.

Output 

* Line 1: The minimum amount of time required for all cows to get under a shelter, presuming they plan their routes optimally. If it not possible for the all the cows to get under a shelter, output "-1".

Sample Input 


3 4
7 2
0 4
2 6
1 2 40
3 2 70
2 3 90
1 3 120

Sample Output 


110

Hint 

OUTPUT DETAILS: 

In 110 time units, two cows from field 1 can get under the shelter in that field, four cows from field 1 can get under the shelter in field 2, and one cow can get to field 3 and join the cows from that field under the shelter in field 3. Although there are other plans that will get all the cows under a shelter, none will do it in fewer than 110 time units.

I used min cost flow in this problem. My algo is:
1. Create a flow network using original graph with capacities infinity. Those fields that have some cows are the sources and the capacity between super source and them is the number of cows in that field. Those fields that have some shelter are the sinks and the capacity between them and super source is the capacity of that shelter. The distance i.e time from super source to sources and from sinks to super sinks are all zero.

2. Use Ford Fulkerson and find augmenting path using Bellman Ford.

I want to know if my algo is correct or not. Any help will be appreciated.

Regards

Sanny

nibbler
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Post by nibbler »

I used similar method, only I used binary search to limit the maximal path length and then maximal flow to see if all the cows can get to shelter. I had a nasty bug so actually won no points on it, but later I had 10 test cases on time, and the rest was TL.
In fact, no one solved this problem completely!

misof
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Post by misof »

The problem I see with using min-cost max-flow: How do you want to compute the answer from the flow? Your task is to minimize the maximum time a cow needs to get to a shelter, not the sum of the times all cows need (IMHO this is what the min-cost flow actually minimizes).

Nibbler's approach is correct, maybe using a fast matching algorithm (Hopcroft-Karp) instead of the flow and efficiently updating the graph between the iterations of the bsearch would help.

Sanny
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Post by Sanny »

Thanks for the reply. Actually I know the binary search method but haven't solved a problem using min cost flow yet. So I thought if I could solve this problem using min cost flow.

Regards
Sanny

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