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[10005],visit[10005],v[10005],ans;
void Prime()//欧拉筛 
{
    for(int i=2;i<=10000;i++)
    {
        if(!visit[i])
        {
            prime[++prime[0]]=i;
        }
        for(int j=1;j<=prime[0]&&i*prime[j]<=10000;j++)
        {
            visit[i*prime[j]]=1;
            if(i%prime[j]==0)break;
        }
    }
}
struct node{
    int a[4],num,cnt;
};
int change(node p)
{
    return p.a[0]*1000+p.a[1]*100+p.a[2]*10+p.a[3];
}
queue<node>q;
int bfs(node p)
{
    while(!q.empty())
    q.pop();
    q.push(p);
    p.num=change(p);
    v[p.num]=1;
    while(!q.empty())
    {
        p=q.front();
        
        if(p.num==y)return p.cnt;
        q.pop();
        for(int i=0;i<4;i++)
        {
            for(int j=0;j<10;j++)
            {
                if(i==0&&j==0)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]=1;
                }
            }
        }
    }
    return -1;
}
int main()
{
    Prime();
    cin>>n;
    while(n--)
    {
        cin>>x>>y;
        for(int i=1000;i<=10000;i++)
        v[i]=0;
        node p;
        int num=x;
        for(int i=3;i>=0;i--)
        {
            p.a[i]=num%10;
            num/=10;
        }
        p.num=change(p);
        p.cnt=0;
        ans=bfs(p);
        if(ans!=-1)cout<<ans<<endl;
        else cout<<"Impossible"<<endl;
    }
    return 0;
}