poj 3126 Prime Path bfs

题目链接：http://poj.org/problem?id=3126

Prime Path

Time Limit: 1000MS		Memory Limit: 65536K
Total Submissions: 32036		Accepted: 17373

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

求出素数在对每个位上的数进行bfs即可

#include<iostream>
#include<queue>
using namespace std;
int n,x,y,prime[],visit[],v[],ans;
void Prime()//欧拉筛
{
    for(int i=;i<=;i++)
    {
        if(!visit[i])
        {
            prime[++prime[]]=i;
        }
        for(int j=;j<=prime[]&&i*prime[j]<=;j++)
        {
            visit[i*prime[j]]=;
            if(i%prime[j]==)break;
        }
    }
}
struct node{
    int a[],num,cnt;
};
int change(node p)
{
    return p.a[]*+p.a[]*+p.a[]*+p.a[];
}
queue<node>q;
int bfs(node p)
{
    while(!q.empty())
    q.pop();
    q.push(p);
    p.num=change(p);
    v[p.num]=;
    while(!q.empty())
    {
        p=q.front();

        if(p.num==y)return p.cnt;
        q.pop();
        for(int i=;i<;i++)
        {
            for(int j=;j<;j++)
            {
                if(i==&&j==)continue;
                node w=p;
                w.a[i]=j;
                w.num=change(w);
                if(!visit[w.num]&&!v[w.num])
                {
                    w.cnt++;
                    q.push(w);
                    v[w.num]=;
                }
            }
        }
    }
    return -;
}
int main()
{
    Prime();
    cin>>n;
    while(n--)
    {
        cin>>x>>y;
        for(int i=;i<=;i++)
        v[i]=;
        node p;
        int num=x;
        for(int i=;i>=;i--)
        {
            p.a[i]=num%;
            num/=;
        }
        p.num=change(p);
        p.cnt=;
        ans=bfs(p);
        if(ans!=-)cout<<ans<<endl;
        else cout<<"Impossible"<<endl;
    }
    return ;
}