学步园

                 万恶的VC6.0，头文件没写竟然也不报错，让我得到两个CompileError……

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

Output

Sample Input

3
1033 8179
1373 8017
1033 1033

Sample Output

6
7
0

#include<iostream>
#include<cstring>
#include<cmath>
using namespace std;

int not_prime[10000]={0};
int queue[10000]={0};
int step[10000]={0};
int visited[10000]={0};
int d[5];//d[i]存第i位数字
int BFS(int num1, int num2)
{
    int i, j;
    int number;
    int temp;
	int m;
    int front=0, rear=1;
    queue[front]=num1;
    visited[num1]=1;//访问过了
    
    while( front!=rear)
    {
        temp=queue[front];
		front++;
        for( j=1; j<=4; j++)//改变第j位数字
        { 
            m=pow(10, 4-j);
			d[j]=temp/m%10;//得到第j位上的数字
			i=(j==1)?1:0;//如果改变第一位数字，则数字从1开始取，其他从0开始
			while( i<=9 )
			{
				if(j==4 &&(i%2==0||i%5==0))//剪枝
				{
					i++;
					continue;
				}
				number=temp+(i-d[j])*m;
				if( number==num2)//搜索到，结束
                    return step[temp]+1;
				if( !visited[number] && !not_prime[number])//若未访问过，且为素数
				{
					queue[rear]=number;
					visited[number]=1;
					step[number]=step[temp]+1;
					rear++;
				}
				i++;
			}
        }
    } 
	return 0;
}
int main()
{
    int n;
    int num1, num2;
	int i, j;
	for(i=2; i<10000; i++)//判断不是素数就赋值为1
		for(j=2; i*j<10000; j++)
			not_prime[i*j]=1;
		cin>>n;
		while( n--)
		{
			memset(visited, 0, sizeof(visited));
			memset(step, 0, sizeof(step));
			cin>>num1>>num2;
			if( num1== num2 )
				cout<<0<<endl;
			else
				cout<<BFS(num1, num2)<<endl;
		}
}