[UVA][bfs] 12101 - Prime Path
Problem G - Prime Path
Time limit: 1 second
The ministers of the cabinet were quite upset by the
message from the Chief of Security stating that they
would all have to change the four-digit room numbers
on their offices.
— It is a matter of security to change such things
every now and then, to keep the enemy in the dark.
— But look, I have chosen my number 1033 for good
reasons. I am the Prime minister, you know!
— I know, so therefore your new number 8179 is also
a prime. You will just have to paste four new
digits over the four old ones on your office door.
— No, it's not that simple. Suppose that I change the
first digit to an 8, then the number will read 8033
which is not a prime!
— I see, being the prime minister you cannot stand
having a non-prime number on your door even for a
few seconds.
— Correct! So I must invent a scheme for going from
1033 to 8179 by a path of prime numbers where
only one digit is changed from one prime to the
next prime.
Now, the minister of finance, who had been eavesdropping,
intervened.
— No unnecessary expenditure, please! I happen to
know that the price of a digit is one pound.
— Hmm, in that case I need a computer program to
minimize the cost. You don't know some very cheap
software gurus, do you?
— In fact, I do. You see, there is this programming
contest going on...
Help the prime minister to find the cheapest prime path between any two given four-digit primes! The first digit must be nonzero, of course. Here is a solution in the case above.
1033
1733
3733
3739
3779
8779
8179
The cost of this solution is 6 pounds. Note that the digit 1
which got pasted over in step 2 can not be reused in the last step
– a new 1 must be purchased.
Input
One line with a positive number: the number of test cases (at most 100). Then for each test case, one line with two numbers separated by a blank. Both numbers are four-digit primes (without leading zeros).Output
One line for each case, either with a number stating the minimal cost or containing the word Impossible.Sample Input
3 1033 8179 1373 8017 1033 1033
Sample Output
6 7 0
在四位質數中，藉由一位一位地替換，問最少步數。
#include <stdio.h>
int p[10000] = {};
void sieve() {
int i, j;
for(i = 2; i < 10000; i++) {
if(p[i] == 0) {
for(j = i+i; j < 10000; j += i)
p[j] = 1;
}
}
}
int main() {
sieve();
int a, b;
scanf("%*d");
while(scanf("%d %d", &a, &b) == 2) {
int Q[9999], Qt, used[10000] = {};
int step[10000] = {}, i, j, k, tn;
char buf[10];
Q[Qt = 0] = a, used[a] = 1;
for(i = 0; i <= Qt && !used[b]; i++) {
tn = Q[i];
for(k = 0; k < 4; k++) {
sprintf(buf, "%d", tn);
for(j = '0'; j <= '9'; j++) {
buf[k] = j;
sscanf(buf, "%d", &a);
if(p[a] == 0 && used[a] == 0 && a >= 1000) {
used[a] = 1, step[a] = step[tn]+1;
Q[++Qt] = a;
}
}
}
}
if(used[b])
printf("%d\n", step[b]);
else
puts("Impossible");
}
return 0;
}