poj 3126 Prime Path BFS

Prime Path

Time Limit: 1000MS		Memory Limit: 65536K
Total Submissions: 6010		Accepted: 3440

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：给两个质数，分别为起点终点。每次可以变动一位为下一个质数，问最少经过多少次变动才能得到终点的质数。

思路：最简单的BFS。

#include<cstdio>
#include<cmath>
#include<cstring>
#include<queue>
using namespace std;
short time[9999];
bool is_prime(int x)
{
	if(x%2==0) return false;
	for(int i=3;i<=(int)(sqrt((double)x)+0.5);i+=2)
	{
		if(x%i==0) return false;
	}
	return true;
}
short bfs(int start,int end)
{
	if(start==end) return 0;
	int now,tmp,a,b,c,d;
	memset(time,-1,sizeof(time));
	time[start]=0;
	queue<short> que;
	que.push(start);
	while(!que.empty())
	{
		now=que.front();que.pop();
		d=now;
		a=d%10;d/=10;
		b=d%10;d/=10;
		c=d%10;d/=10;
		for(int i=1;i<=9;i++)
		{
			tmp=i*1000+c*100+b*10+a;
			if(tmp==end) return time[now]+1;
			if( time[tmp]==-1 && is_prime(tmp) )
			{
				que.push(tmp);
				time[tmp]=time[now]+1;
			}
		}
		for(int i=0;i<=9;i++)
		{
			tmp=d*1000+i*100+b*10+a;
			if(tmp==end) return time[now]+1;
			if( time[tmp]==-1 && is_prime(tmp) )
			{
				que.push(tmp);
				time[tmp]=time[now]+1;
			}
			tmp=d*1000+c*100+i*10+a;
			if(tmp==end) return time[now]+1;
			if( time[tmp]==-1 && is_prime(tmp) )
			{
				que.push(tmp);
				time[tmp]=time[now]+1;
			}
			tmp=d*1000+c*100+b*10+i;
			if(tmp==end) return time[now]+1;
			if( time[tmp]==-1 && is_prime(tmp) )
			{
				que.push(tmp);
				time[tmp]=time[now]+1;
			}
		}
	}
}
int main()
{
	int T;scanf("%d",&T);
	int start,end;
	while(T--)
	{
		scanf("%d%d",&start,&end);
		printf("%d\n",bfs(start,end));
	}
	return 0;
}