11377 - Airport Setup

[Giorgi]

this problem makes me crazy...
I can't understand why simple soltion with BFS did not get AC.
is there any tricky case or I can't understand problem statement well?

[rio]

Does your code handle the case x==y ?

-----
Rio

[Giorgi]

Does your code handle the case x==y ?

Yes

here is my code, maybe I have very stupid mistake , but I can't find it...

Code: Select all

I was Code :)

[rio]

Does your code handle the case x==y ?

Yes

No. Your code is not handling it correctly.

For input:

Code: Select all

1
2 1 1
1
1 2
1
2 2

You must output 0.
-----
Rio

[Giorgi]

thank you I got AC. very stupid mistake

[joshi13]

Hi, i'm wondering how would be the graph you're working with, how to construct it. It would be very helpful.
Also please post some I/O cases.

Thanks for replies.

[srrajesh]

I first implemented BFS, but that gave WA, even after considering x == y case.
So I implemented Djkstra noting that if x and y don't have an airport then the cost of path between them will be more than, if any one of them have airport which in turn will be more when, both have airport in it.
Is this approach correct?
I am getting continuous WAs
This is my first implementation of Djkstra, so I think there might be some problem with the implementation.

Code: Select all

Code removed after getting AC

Thanks in advance for your help.

[mf]

So I implemented Djkstra noting that if x and y don't have an airport then the cost of path between them will be more than, if any one of them have airport which in turn will be more when, both have airport in it.
Is this approach correct?

As you implemented it, no.
At very least you should assign a cost of 0 to edges with two airports, or else your solution would always prefer shorter paths to longer ones, even if longer paths would require to build less airports.

Here's one correct approach: let cost of directed edge a->b be 1 if b doesn't have an airport, and 0 otherwise; each undirected edge (a, b) corresponds to two symmetric directed edges: a->b and b->a.
Then the answer is length of shortest path between x and y, plus 1 if x doesn't have an airport.

[srrajesh]

Here's one correct approach: let cost of directed edge a->b be 1 if b doesn't have an airport, and 0 otherwise; each undirected edge (a, b) corresponds to two symmetric directed edges: a->b and b->a.
Then the answer is length of shortest path between x and y, plus 1 if x doesn't have an airport.

I implemented as you said now, but still WA.
In my previous post, I had a very serious bug!
I had declared "arr" adjacency matrix as bool, so no effect of assigining it any non-zero value to it!
Indeed, it is a very stupid mistake.
I corrected it as char and resubmitted it with the idea you said, but still I am getting WA.
In previous posts, people say that they solved with BFS. If so can anyone give me an idea.

Code: Select all

Code removed after getting AC

[mf]

*Always* remember to test your code on sample I/O, especially before posting it here.
It prints extra newlines.

In previous posts, people say that they solved with BFS. If so can anyone give me an idea.

You can find shortest paths in graphs with edge weights 0 and 1 with a modification of BFS: instead of a queue use deque, when you process an edge of cost 1, insert the new node at the back (as in BFS), else at the front of the deque.

[srrajesh]

*Always* remember to test your code on sample I/O, especially before posting it here.
It prints extra newlines.

I got AC now
Thanks a million!
I added some debugging statements, and then I deleted them, but I forgot to remove the newline I had given as a part of that.
I should have tested by redirecting the o/p to a file, before posting. Thanks a million for your help.

You can find shortest paths in graphs with edge weights 0 and 1 with a modification of BFS: instead of a queue use deque, when you process an edge of cost 1, insert the new node at the back (as in BFS), else at the front of the deque.

Thanks for the idea.
If I understand correctly, I think this concept applies when the weights are 0 and any other value >= 1.

[hsn]

hello everyone i really need some help here
i have done this problem by using floyd algorithm
i don't know if this is the correct approch or not.
when i give 2000 cities it take alot of time for the code to finish so it is very slow.
what is the best way to solve this question and why??????????????
plzzzzzz help me i really need to solve this question.

thank in advance

[sapnil]

I use simple dijkstra, But till WR
Here my code

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// Code removed

Thanks
Keep Posting
Sapnil

[kbr_iut]

thanku mf.
I got AC using dijsktra with ur explanation.

[tanmoy]

can any one help me about this problem plz

Code: Select all

#include<stdio.h>
#include<memory.h>
#include<iostream>
#include<queue>
using namespace std;
#define MAX 2002
int adj[MAX][MAX];
int P[MAX+3],O[MAX];
bool T[MAX];
int n,s,d,counter,v=0;

void BFS (void){
	int x,i,c=0;
	bool flag=false;
	fill(P,P+n+2,0);
	queue<int> Q;
	Q.push(s);
	P[s]=1;
	O[s]=0;
	while(!Q.empty()&&Q.back()!=d){
		x=Q.front();
		Q.pop();
		for(i=1;i<=n;i++){
			if(adj[x][i]==1 && P[i]==0){
				P[i]=1;
				Q.push(i);
				O[i]=x;
				if(Q.back()==d){flag=true;break;}
			}
		}
	}
	v++;
	printf("Case %d:\n",v);
	if(flag==false && s!=d ){printf("-1\n");return ;}
	else if(s==d){
			printf("0\n");
			return;
	}
		counter=0,i=1;
		if(T[d]==false)counter++;
		x=d;
		while(1){
			x=O[x];
			if(T[x]==false)counter++;
			if(x==s) break;
			x=O[x];
			if(T[x]==false)counter++;
			if(x==s )break;
		}
		printf("%d\n",counter);


}






int main(){
	int X,N,M,K,i,g;
	scanf("%d",&X);
	while(X--){
		memset(T,false,sizeof(T));
		scanf("%d %d %d",&N,&M,&K);
		n=N;
		for(i=1;i<=N;i++){
			memset(adj[i],0,sizeof(int)*N);
		}
		for(i=0;i<K;i++){
			scanf("%d",&g);
			T[g]=true;
		}
		for(i=0;i<M;i++){
			scanf("%d %d",&g,&K);
			adj[g][K]=1;
			adj[K][g]=1;
		}
		scanf("%d",&K);
		while(K--){
			scanf("%d %d",&s,&d);
			BFS();
		}
	}
	return 0;
}