OJ Board

tat tvam asi wrote: 10 packet 1->4 (10 time unit)
10 packet 1->2->1->4 (50 time unit) ?

Aren't you using then edges 1-2 and 1-4 two times?

4 4
1 3 3
3 4 4
1 2 2
2 4 5
20 1



You cannot send 20 packets through this network, if edge capacity is 1 - you'll send 1 packet through 1-3-4 and one through 1-2-4, but no more.


4 5
1 4 1
1 3 3
3 4 4
1 2 2
2 4 5
20 10

is 80?

Sending ten packets through 1-4, which takes 10 * 1 = 10 ...

[cpp]
for(i=1;i<n;i++){
do{

q=pie[q];
if(P[q+1]==T )
q=q+1;
if(q==m)
printf("pattern occurs with shift %d",i-m);
q=pie[q];

check++;

}while(q>0 && P[q+1]!=T );
[/cpp]

I'm not sure i did understood your code (i use Pascal), but, as i see
1. If, at the beginning of a loop, q=m ...

Hello,
if two events happen at the same time, in what order should they be printed?

Best regards,

Furippo

Hello,
I've been trying for some time to solve this problem, and i have no idea, why i get WA all the time. Is there any trick in this task? I solve it with DP, for each k and l, i find the maximum number of ridings, if you can use k avenues to divide, and the last to the north would be l avenue ...

Hello,
i've got a problem with this task too.
this is my alghoritm:
the maximal word length is the length of the sequence, which we have to find in all D0L words. If i add new word to a queue, and it's longer than maximal length, i add word consisting of m (m=maximal length) first characters of ...

The problem is not here. I would like to know, what problems may occur, when i'm adding two big numbers as strings in GNU Pascal, becouse - like i said - our outputs for every possible input were the same (so the problem is not in the way of adding them).

Hello,
i've got a little problem with this task. I got WA, but i've checked my output with my friend's program, and in all tests we've got the same outputs, more: our output files were completely the same! My friend's program got AC, of course.

More detail:

N - number of tooses, K - length of sequence. Two possible situations:
1. Sequence of length K appeared after N-1 tooses - T(N-1,K)
2. The last K tosses are heads, N-K toss was tail (otherwise, if tosses through N-K to N-1 were heads, the sequence of length K would appear after N-1 ...