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Posted: Mon Aug 23, 2004 6:27 am
by sohel
Accepted.

I modified my bfs() and got it AC in 0.04 seconds.
Larry wrote: Use dynamic programming by first finding a recurrence..
Can someone give some detail about this recurrence relation. I think most of the people used Larry's method.

Posted: Mon Aug 23, 2004 3:59 pm
by tep
I use memoization / dp approach. Here is some clue

memo[v][d]
can i reach the destination if i'm vertex no v and i have d days left.

I'm sure you can figure out the rest

regards,
stephanus

10681 Teobaldo's Trip

Posted: Thu Sep 01, 2005 4:16 am
by daveon
Here is a tricky case.

INPUT:

Code: Select all

3 2
1 2
2 3
2 2 0
OUTPUT:

Code: Select all

Yes, Teobaldo can travel.

10681 Problem..

Posted: Fri Nov 25, 2005 6:30 am
by Rocky
I Get WA For This Problem....
Can Any One Give Me Some Tricky Case.....

Thank's In Advance
Rocky

Posted: Mon Dec 26, 2005 12:50 am
by daveon

Thank's..

Posted: Sun Jan 01, 2006 7:15 am
by Rocky
Thank's....

Rocky

10681 Teobaldo's Trip again.. Tell me what's wrong..!!

Posted: Thu Jan 05, 2006 6:01 pm
by helloneo
if Teobaldo can go j city on i-th day, i check dp[j] = i
it's kind of straightforward.. but getting WA..
plz tell me what's wrong..


Code: Select all

code removed

Try this input

Posted: Sun Feb 19, 2006 6:29 pm
by medv
Try this input:

2 1
1 2
1 2 3

your program gives NO, but must be YES.
The trip is: 1 - 2 - 1 - 2

Posted: Sat Apr 01, 2006 9:55 am
by asif_rahman0
i m not expert in DP
so i think u can use easily matrix multiplication for finding link between vertices. which is O(n^3).
then use another loop for days....then it would be O(n^4) by some extra checking. get it aceepted within 1.4/1.3

hope it helpssssssss. :)

N.B: dont use map[days][j] = map[days][j] | ( map[days][k]&map[days][k][j])

use IF condition for faster Running Time

Posted: Fri Aug 04, 2006 11:31 pm
by Darko
I guess I don't understand this problem...
What would be the output for

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2 1
1 2
1 2 4
I think it's "No", because he can't make that trip in exactly 4 days?

Posted: Sat Aug 05, 2006 12:02 am
by mf
That's right. Output is "No" if and only if there is no trip of length exactly D days.

Re: 10681 - Teobaldo's Trip

Posted: Tue Jun 07, 2011 10:19 pm
by Shafaet_du
I solved using dp but i am interested to know the matrix exponent solution. I am failing to raise the power to 200 as the elements are becoming too large. Do we need to use some kind of modular arithmetic?

Re: 10681 - Teobaldo's Trip

Posted: Wed Jul 25, 2012 10:49 pm
by pranon
Code being judged ``WA" with me having no idea why. Any help in pointing out would be much appreciated.

Code: Select all

[code]
#include <stdio.h>
#include <string.h>

char mat[105][105], adj[105][105], temp[105][105], city[105];
char ytic[100000];


int main()
{
    int c, l, j, i, a, b, s, e, d, t, k;
    while(scanf("%d %d", &c, &l)==2 && (c || l))
    {
        j=0;
        while(l--)
        {
            scanf("%d %d", &a, &b);
            if(ytic[a]==0)
            {
                city[j]=a;
                ytic[a]=j++;
            }
            if(ytic[b]==0)
            {
                city[j]=b;
                ytic[b]=j++;
            }
            adj[ytic[a]][ytic[b]]=adj[ytic[b]][ytic[a]]=mat[ytic[a]][ytic[b]]=mat[ytic[b]][ytic[a]]=1;
        }
        scanf("%d %d %d", &s, &e, &d);
        t=1;
        s=ytic[s];
        e=ytic[e];
        while((t<<1)<=d)
        {
            for(i=0; i<c; i++)
                for(j=0; j<c; j++)
                    for(k=0; k<c; k++)
                        if(mat[i][k]>0 && mat[k][j]>0)
                            temp[i][j]=1;
            for(i=0; i<c; i++)
                for(j=0; j<c; j++)
                    mat[i][j]=temp[i][j];
            t<<=1;
        }
        while(t<d)
        {
            for(i=0; i<c; i++)
                for(j=0; j<c; j++)
                    for(k=0; k<c; k++)
                        if(mat[i][k]>0 && adj[k][j]>0)
                            temp[i][j]=1;
            for(i=0; i<c; i++)
                for(j=0; j<c; j++)
                    mat[i][j]=temp[i][j];
            t++;
        }
        (mat[s][e]>0 ||(s==e && d==0))?printf("Yes, Teobaldo can travel.\n"):printf("No, Teobaldo can not travel.\n");
        for(i=0; i<c; i++)
        {
            memset(mat[i], 0, c*sizeof(char));
            memset(adj[i], 0, c*sizeof(char));
        }
        memset(city, 0, c*sizeof(char));
    }
    return 0;
}

[/code]
Thanks in advance. :(

Re: 10681 - Teobaldo's Trip

Posted: Thu Feb 14, 2013 12:29 am
by mostafiz93
I'm getting WA in this problem.

My algo is : first make adjacency matrix with the given liunks. then journey is possible if [adj matrix]^d is nonzero.

i've implemented this with matrix exponentiation.

Can anybody help me?

my code is here:

Code: Select all

removed after AC

Re: 10681 - Teobaldo's Trip

Posted: Thu Jul 24, 2014 7:54 am
by sdipu
When you are solving this problem using matrix exponentiation, don't forget to use modulo operation.
Try this-

input:

Code: Select all

3 2
1 2
2 3
3 1 200

5 7
1 5
2 4
3 5
1 3
2 4
3 2
2 5
3 4 200

0 0
output:

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Yes, Teobaldo can travel.
Yes, Teobaldo can travel.