POJ3126 - Prime Path

                                                                                                                 
Prime Path

Time Limit:1000MS    Memory Limit:65536KB    64bit IO Format:%lld
& %llu

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.

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

这道题目的大意就是给出两个数字，都是素数，要求求前面那个数字转化为后面那个数字的步数。转化规则是，每次只能变当前数字的某一位数字，且生成一个数字后则个数字仍是素数。

这道题看得出就是个广搜的题目，合理运用BFS就能解出来。考虑到1000到10000的素数只有1061个，时间复杂度可以算作是100*1000*1000，每次都遍历一次这些素数看看是不是有一个数字不同其他相同

#include<cstdio>
#include<queue>
#include<cmath>
#include<cstring>
using namespace std;
int prime[10000],k,a,b;
bool mark[10000],flag;

struct node
{
int x;
int Step_Count;
};

bool Check(int x,int y)
{
int cout=0;
while(x>0)
{
if(x%10==y%10)
cout++;
x/=10;
y/=10;
}
if(cout==3)
return 1;
else
return 0;
}

void bfs()
{
node now,next;
queue <node> q;
now.x=a;
now.Step_Count=0;
q.push(now);
while(!q.empty())
{
now=q.front();
if(now.x==b)
{
printf("%d\n",now.Step_Count);
flag=1;
return;
}
for(int i=0;i<k;i++)
{
if(!mark[i] && Check(now.x,prime[i]))
{
next.x=prime[i];
next.Step_Count=now.Step_Count+1;
q.push(next);
mark[i]=1;
}
}
q.pop();
}
}

int main()
{
int i,j,t,T;
for(i=1000;i<10000;i++)
{
flag=1;
t=sqrt(i);
for(j=2;j<=t;j++)
{
if(i%j==0)
{
flag=0;
break;
}
}
if(flag)
prime[k++]=i;
}
scanf("%d",&T);
while(T--)
{
flag=0;
scanf("%d%d",&a,&b);
memset(mark,0,sizeof(mark));
bfs();
if(flag==0)
printf("0\n");
}
return 0;
}