POJ 3126 Prime Path （BFS）

L - Prime Path
Time Limit:1000MS     Memory Limit:65536KB     64bit IO Format:%lld
& %llu
Submit Status Practice POJ
3126

Appoint description: 
System Crawler  (Jul 30, 2016 2:58:14 PM)

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

题意：给定两个素数a b，求a变幻到b需要几步
                并且变幻时只有一个数字不同，并且变换后的数是素数

#include<stdio.h>
#include<string.h>
#include<queue>
#include<algorithm>
#define MAX 10000
using namespace std;
bool prime[MAX];
void init()
{    //对素数打表
int i,j;
for(i=1000;i<=MAX;i++)
{
for(j=2;j<i;j++)
if(i%j==0)
{
prime[i]=false;
break;
}
if(j==i)
prime[i]=true;
}
}
int bfs(int first,int last)
{
int count[MAX],t[5],i,j,k,l,v,temp;
bool vis[MAX];
memset(count,0,sizeof(count));
memset(vis,false,sizeof(vis));
queue<int>q;
q.push(first);
vis[first]=true;
while(!q.empty())
{
v=q.front();
q.pop();
t[0]=v/1000%10;
t[1]=v/100%10;
t[2]=v/10%10;
t[3]=v%10;
for(i=0;i<4;i++)
{
temp=t[i];
for(j=0;j<10;j++)
if(j!=temp)
{
t[i]=j;
k=t[0]*1000+t[1]*100+t[2]*10+t[3];
if(!vis[k]&&prime[k])
{
count[k]=count[v]+1;
vis[k]=true;
q.push(k);
}
if(k==last)
return count[k];
}
t[i]=temp;
}
if(v==last)
return count[v];
}
return -1;
}
int main()
{
int t,a,b;
scanf("%d",&t);
init();//这个不能放到while里面
while(t--)
{
scanf("%d%d",&a,&b);
int ans=bfs(a,b);
if(ans==-1)
printf("Impossible\n");
else
printf("%d\n",ans);
}
return 0;
}