Prime Path POJ - 3126 BFS

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

将一个四位数逐位换成另一个四位数，不能有前导零，变换过程中的数也必须是质数。

先将质数打表，再BFS搜索，换掉一位数还是质数则加入下一层

#include <string.h>
#include <iostream>
#include <queue>
#include <stdio.h>
using namespace std;
int pr[10001];
int vis[10001];
int dis[10001];
int mo[4]={1,10,100,1000};
void table(){
for(int i=2;i<=10000;i++){
if(pr[i]){
for(int j=2*i;j<=10000;j+=i){
if(pr[j]){
pr[j]=0;
}
}
}
}

}

int bfs(int a,int b){
if(a==b){
return 0;
}
queue<int> q;
dis[a]=0;vis[a]=1;
q.push(a);
int w[4];
int x,y,t,s;
while(!q.empty()){
x=q.front();
//cout<<x<<' '<<dis[x]<<endl;
q.pop();
if(x==b){
return dis[b];
}
y=dis[x];
for(int i=0;i<4;i++){
w[i]=x%10;
x/=10;
}
for(int i=0;i<4;i++){
for(int j=1;j<=9;j++){

t=(w[i]+j)%10;
if(i==3&&t==0)
continue;
s=0;
for(int k=0;k<4;k++){
if(k==i){
s+=t*mo[k];
}else{
s+=w[k]*mo[k];
}
}
if(pr[s]&&!vis[s]){
vis[s]=1;
dis[s]=y+1;
q.push(s);

}

}
}
}
return -1;
}

int main()
{
int T,a,b,k;
scanf("%d",&T);
memset(pr,1,sizeof(pr));
table();
while(T--){
memset(vis,0,sizeof(vis));
scanf("%d %d",&a,&b);
k=bfs(a,b);
if(k==-1){
printf("Impossible\n");
}else{
printf("%d\n",k);
}
}

return 0;
}