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804 - Petri Net Simulation
Posted: Mon Aug 22, 2005 11:32 pm
by Jan
I have solved the problem but getting WA.
Can anyone tell me what should be the output for the following input set...
Input:
Code: Select all
2
1 0
2
-1 2 0
-2 1 0
100
3
3 0 0
3
-1 2 0
-2 -2 3 0
-3 1 0
100
3
1 0 0
3
-1 2 3 0
-2 1 0
-3 1 0
1
3
3 0 0
3
-1 2 0
-2 -2 3 0
-3 1 0
9
3
3 0 0
3
-1 2 0
-2 -2 3 0
-3 1 0
10
3
3 0 0
3
-1 2 0
-2 -2 3 0
-3 1 0
8
0
My code returns
Output:
Code: Select all
Case 1: still live after 100 transitions
Places with tokens: 1 (1)
Case 2: dead after 9 transitions
Places with tokens: 2 (1)
Case 3: still live after 1 transitions
Places with tokens: 2 (1) 3 (1)
Case 4: still live after 9 transitions
Places with tokens: 2 (1)
Case 5: dead after 9 transitions
Places with tokens: 2 (1)
Case 6: still live after 8 transitions
Places with tokens: 1 (1)
Thanks in advance.
Posted: Tue Aug 23, 2005 12:24 pm
by Jan
Finally got accepted.
The output given is correct...
How to make it fast?
Posted: Sun Feb 12, 2006 9:57 am
by imnew
People are solving this within a pretty small time limit........................I could get no better idea than brute force simulation.........Any idea?
Posted: Tue Mar 28, 2006 10:24 am
by sclo
I keep getting TLE for this problem too.
Posted: Fri Mar 31, 2006 5:23 am
by sclo
The 10 sec time limit might be too tight for this problem.
Posted: Fri May 26, 2006 9:11 pm
by lpereira
Brute force will do it. Remember that you can greedly get the first active transition you find. This should speed up. Also, if one of the imputs do not suffice, you can also discard that transition, since it's surely not active.
Posted: Mon Jul 10, 2006 10:05 am
by p137
My program returns exactly the same output, but I still get "wrong answer". Has anybody an idea, what kind of reason this could have? (Is it important whether I print a blank line before/after output?)
@Jan: What did you do to get accepted?
Thanks in advance!
p137
Posted: Wed Oct 25, 2006 12:19 am
by Jan
p137 wrote:@Jan: What did you do to get accepted?
I just simulate the process efficiently. May be you have missed this line
The input data will be selected to guarantee the uniqueness of the correct output displays.
Hope it helps
.
Re: 804 - Petri Net Simulation
Posted: Fri Nov 21, 2008 11:28 pm
by lucaskt
I can't find any test cases that make this fail, but am still getting WA, any ideas?
Code: Select all
#include <stdio.h>
int main() {
int mi[120][120];
int mc[120][120];
int s[120];
int c, d, f, g, n, t;
int i, j, k;
scanf(" %d", &n);
g = 1;
while (n != 0) {
if (g > 1) putchar('\n');
for (i = 0; i < n; i++) {
scanf(" %d", &s[i]);
}
scanf(" %d", &t);
for (i = 0; i < n; i++) {
for (j = 0; j < t; j++) {
mc[i][j] = mi[i][j] = 0;
}
}
for (i = 0; i < t; i++) {
scanf("%d", &k);
while (k != 0) {
if (k < 0) {
mi[-k - 1][i]++;
mc[-k - 1][i]--;
} else {
mc[k - 1][i]++;
}
scanf("%d", &k);
}
}
scanf(" %d", &f);
c = 0;
while (c < f) {
d = 0;
for (i = 0; i < t; i++) {
for (j = 0; j < n; j++) {
if (s[j] < mi[j][i]) {
break;
}
}
if (j == n) {
for (j = 0; j < n; j++) {
s[j] += mc[j][i];
}
c++;
d = 1;
if (c == f) goto end;
}
}
if (!d) {
break;
}
}
end:
if (!d) {
printf("Case %d: dead after %d transitions\n", g++, c);
} else {
printf("Case %d: still live after %d transitions\n", g++, c);
}
printf("Places with tokens:");
for (i = 0; i < n; i++) {
if (s[i] > 0) {
printf(" %d (%d)", i + 1, s[i]);
}
}
putchar('\n');
scanf(" %d", &n);
}
return 0;
}
I've tried the set Jan sent, and it got the same answers. I also tried this (appended to Jan's set):
Code: Select all
2
2 0
2
-1 2 2 0
-2 -2 -2 1 0
10
2
2 0
2
-1 2 2 0
-2 -2 -2 1 0
6
expecting:
Code: Select all
Case 7: dead after 6 transitions
Places with tokens: 2 (2)
Case 8: still live after 6 transitions
Places with tokens: 2 (2)
and it worked, but the judge still gives me WA.
Thanks in advance,
Lucas
Re: 804 - Petri Net Simulation
Posted: Mon Oct 23, 2017 11:22 pm
by jurajz
I had many WA's because I didn't assume, that NT (number of transitions) can be zero. After this assumption, I had AC.
Sample input:
Code: Select all
2
1 0
2
-1 2 0
-2 1 0
100
2
0 0
2
-1 2 0
-2 1 0
0
3
3 0 0
3
-1 2 0
-2 -2 3 0
-3 1 0
100
2
0 0
2
-1 2 0
-2 1 0
0
0
Correct output:
Code: Select all
Case 1: still live after 100 transitions
Places with tokens: 1 (1)
Case 2: still live after 0 transitions
Places with tokens:
Case 3: dead after 9 transitions
Places with tokens: 2 (1)
Case 4: still live after 0 transitions
Places with tokens: