Prime Path
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
解题思路：打一个素数表来判断每次变化后的数是否为素数，通过bfs搜索寻找最短变化时间（即最小使用英镑），关键在于操作的模拟，即每次位数上的变化，我们应利用双层for循环来解决。
AC代码：
#include<iostream>
#include<algorithm>
#include<queue>
#include<cstdlib>
#include<cmath>
#include<memory.h>
#include<cstdio>
using namespace std;
int t,a,b;
const int maxn=10000;
int result[maxn];
bool isprime[maxn];
bool prime(int n) //判断是否是素数
{
int t=sqrtl(n);
for(int i=2;i<=t;i++)
{
if(n%i==0)return false;
}
return true;
}
int change(int n,int i) //将第i位上的数置为0；
{
char str[6]={0};
sprintf(str,"%d",n);
str[i]='0';
sscanf(str,"%d",&n);
return n;
}
bool bfs(int a,int b)
{
queue<int> Q;
int temp,head,t;
Q.push(a);
result[a]=1;
while(!Q.empty())
{
head=Q.front();
if(head==b)return true;
Q.pop();
t=1000;
for(int i=0;i<4;i++)
{
temp=change(head,i);
int temp1=temp;
for(int j=0;j<10;j++){
temp=temp1+j*t;
if(isprime[temp]&&result[temp]==0){
Q.push(temp);
result[temp]=result[head]+1;
}
}
t/=10;
}
}
return false;
}
int main()
{
for(int i=1000;i<=maxn;i++)
isprime[i]=prime(i);
while(cin>>t)
{
for(int i=0;i<t;i++)
{
memset(result,0,sizeof(result));
cin>>a>>b;
if(bfs(a,b))cout<<result[b]-1<<endl;
else cout<<"Impossible"<<endl;
}
}
return 0;
}