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
题意：
给出n和m，要求把n变为m，每次只能变一位上的数字，且每次中间过程的四位数也必须是素数，求最少变化次数。
解题思路：
BFS。遍历四位数字上的每一位即可。
AC代码：
#include<cstdio>
#include<cstring>
#include<queue>
using namespace std;
int n,m;
int book[10010];
struct node
{
int x,step;
};
node getnode(int x,int step)
{
node q;
q.x=x;
q.step=step;
return q;
}
int isp(int n)
{
if(n<=1) return 0;
else
{
for(int i=2;i*i<=n;i++)
{
if(n%i==0) return 0;
}
}
return 1;
}
void bfs(int x,int step)
{
queue<node> q;
q.push(getnode(x,step));
while(!q.empty())
{
if(q.front().x==m)
{
printf("%d\n",q.front().step);
return;
}
int a=q.front().x/1000;
int b=q.front().x/100%10;
int c=q.front().x%100/10;
int d=q.front().x%10;
//printf("a===%d b===%d c===%d d===%d\n",a,b,c,d);
for(int i=0;i<=9;i++)
{
if(i==a) continue;
else
{
int tx=i*1000+b*100+c*10+d;
if(tx>=1000&&isp(tx)&&book[tx]==0)
{
book[tx]=1;
q.push(getnode(tx,q.front().step+1));
}
}
}
for(int i=0;i<=9;i++)
{
if(i==b) continue;
else
{
int tx=a*1000+i*100+c*10+d;
if(tx>=1000&&isp(tx)&&book[tx]==0)
{
book[tx]=1;
q.push(getnode(tx,q.front().step+1));
}
}
}
for(int i=0;i<=9;i++)
{
if(i==c) continue;
else
{
int tx=a*1000+b*100+i*10+d;
if(tx>=1000&&isp(tx)&&book[tx]==0)
{
book[tx]=1;
q.push(getnode(tx,q.front().step+1));
}
}
}
for(int i=0;i<=9;i++)
{
if(i==d) continue;
else
{
int tx=a*1000+b*100+c*10+i;
if(tx>=1000&&isp(tx)&&book[tx]==0)
{
book[tx]=1;
q.push(getnode(tx,q.front().step+1));
}
}
}
q.pop();
}
printf("Impossible\n");
}
int main()
{
int t;
scanf("%d",&t);
while(t--)
{
scanf("%d%d",&n,&m);
memset(book,0,sizeof(book));
book[n]=1;
bfs(n,0);
}
return 0;
}