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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:
OUTPUT:
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..
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
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:
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:
Code: Select all
Yes, Teobaldo can travel.
Yes, Teobaldo can travel.