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
题意如下：
就是给出两个素数a，b， 每次操作只允许将a中的一个数字变换，问最少经过多少次才能将a变换成b。
当时一看到最短，
首先想到的是bfs， 然后发现可以将4位素数存储起来， 用于搜索。 。
代码如下：
#include <cstdio>
#include <cstring>
#include <iostream>
#include <algorithm>
#include <queue>
#include <cmath>
using namespace std;
int n;
int a,b;
int is_pr[10005];
int snum=0;
int vis[10005];
struct card
{
int data;
int num;
};
int bfs()
{
if(a==b)
return 0;
vis[a]=1;
queue<card>q;
card now,next;
now.data=a; now.num=0;
q.push(now);
while (!q.empty())
{
now=q.front();
q.pop();
for (int i=0;i<snum;i++)
{
int ci=0;
if(!vis[is_pr[i]])
{
next.data=is_pr[i];
next.num=now.num+1;
int ttemp1=next.data;
int ttemp2=now.data;
for (int j=0;j<4;j++)
{
int aa=ttemp1%10;
int bb=ttemp2%10;
ttemp1/=10;
ttemp2/=10;
if(aa==bb)
{
ci++;
}
}
if(ci==3)
{
//printf("%d %d %d\n",is_pr[i],now.data,now.num);
if(next.data==b)
return next.num;
q.push(next);
vis[next.data]=1;
}
}
}
}
}
int isPrime(int n)
{
if(n<=1)
return 0;
if(n==2||n==3)
return 1;
if(n%6!=5&&n%6!=1)
return 0;
for(int i=5;i<=sqrt(n);i++)
if(n%i==0||n%(i+2)==0)
return 0;
return 1;
}
int main()
{
scanf("%d",&n);
for (int i=1000;i<=9999;i++)
{
if(isPrime(i))
{
is_pr[snum++]=i;
//printf("%d\n",is_pr[snum-1]);
}
}
while (n--)
{
scanf("%d%d",&a,&b);
memset (vis,0,sizeof(vis));
printf("%d\n",bfs());
}
return 0;
}