poj3126 bfs

http://poj.org/problem?id=3126

Prime Path

Time Limit: 1000MS		Memory Limit: 65536K
Total Submissions: 14460		Accepted: 8159

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，将第一个质数变为第二个质数，过程中，每变化一次，只能变化其中一位数，而且每次变化得到的必须是质数

1033→1733→3733→3739→3779→8779→8179；；；；

简单的bfs下面给出代码

#include <iostream>
#include <string.h>
#include <queue>
#include <math.h>
using namespace std;
bool is_prime[10010];
int tot;
bool f[10005];
int d[10005];
int dx[4]= {1,10,100,1000};
void get_prime()
{
    int i,j;
    for(i=1000; i<=10000; i++)
    {
        for(j=2; j<=(int)sqrt((double)i); j++)///||j*j<=i
        {
            if(i%j == 0)
            {
                is_prime[i]=false;
                break;
            }
            is_prime[i] = true;
        }
    }
}
int jianwei(int m,int i)
{
    int k;
    if(i==0)
        return m/10*10;
    if(i==1)
    {
        k=m%10;
        return m/100*100+k;
    }
    if(i==2)
    {
        k=m%100;
        return m/1000*1000+k;
    }
    if(i==3)
        return m%1000;
}
///bfd
int bfs(int m,int n)
{
    queue<int>q;
    q.push(m);
    memset(f,0,sizeof(f));
    f[m]=1;
    d[m]=0;
    while(q.size())
    {
        int p=q.front();
        q.pop();
        f[p]=1;
        for(int i=0; i<4; i++)
        {
            for(int j=0; j<10; j++)
            {
                int low=jianwei(p,i);
                int k=dx[i]*j+low;///遍历0——9，将该位数字遍历一遍，
                if(!f[k]&&is_prime[k])///该数没有用过，并且是质数
                {
                    q.push(k);///将该数加入队列
                    d[k]=d[p]+1;///距离 +1
                    f[k]=1;///标记为已经使用过
                }
                if(k==n)
                    break;

            }
        }
    }
    return  d[n];
}
int main()
{
    get_prime();
    int m,n,ans,t;
    cin>>t;
    while(t--)
    {
        cin>>m>>n;
        ans= bfs(m,n);
        cout<<ans<<endl;
    }
    return 0;
}