题目链接：http://poj.org/problem?id=3126
Description
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.
1033The 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.
1733
3733
3739
3779
8779
8179
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
题意：
给出两个四位数的素数a,b，求从a变到b最少要花几步？每一步只能将a中的一位改变，且改变后的数也要是素数。
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思路：
枚举搜索，我不知道怎么得到的，但我可以一直搜~。千位，百位，十位，个位的数依次替换，依次深入搜索。
注意：
以前总喜欢把标记路径的Vir放在出队时，其实应该放在入队的时候，这样就不会有一样的数据入队，节省了不少时间和空间。
AC代码:
| Time | 63ms |
|---|---|
| Memory | 608kB |
| Length | 1363 |
#include <cstdio>
#include <cmath>
#include <queue>
#include <cstring>
using namespace std;
struct Q
{
int x[4];
int s;
}no,de;
queue<Q> iq;
int n,a,b,ans,t[4];
int vir[10008];
bool isPrime( int x )
{
for( int i=2; i<=sqrt(x); i++ )
if( x%i == 0 )
return false;
return true;
}
int wirte( int* a)
{
return a[0]*1000+a[1]*100+a[2]*10+a[3];
}
void bfs( )
{
while( !iq.empty() )
iq.pop();
ans=0;
memset(vir,0,sizeof(vir));
no.s=0;
for( int i=0; i<4; i++ )
{
no.x[3-i]=a%10;
a/=10;
}
iq.push(no);
vir[wirte(no.x)] = 1 ;
while( !iq.empty() )
{
no = iq.front();
iq.pop();
if( wirte(no.x)==b )
{
printf("%d\n",no.s);
return ;
}
for( int i=0; i<4; i++ )
{
de=no;
for( int j=0; j<=9; j++ )
{
if( i==0 && j==0 ) continue;
de.x[i] = j ;
if( !vir[wirte(de.x)] && isPrime(wirte(de.x)))
{
vir[wirte(de.x)] = 1 ;
de.s=no.s+1;
iq.push(de);
}
}
}
}
printf("Impossible\n");
}
int main()
{
scanf("%d",&n);
while( n-- )
{
scanf("%d%d",&a,&b);
bfs();
}
return 0;
}