CD0J/POJ 851/3126 方老师与素数/Prime Path BFS

Prime Path

Time Limit: 1000MS		Memory Limit: 65536K
Total Submissions: 9982		Accepted: 5724

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

#include <cstdio>
#include <cmath>
#include <cstring>
#include <ctime>
#include <iostream>
#include <algorithm>
#include <set>
#include <vector>
#include <sstream>
#include <queue>
#include <typeinfo>
#include <fstream>
typedef long long ll;
using namespace std;
//freopen("D.in","r",stdin);
//freopen("D.out","w",stdout);
#define sspeed ios_base::sync_with_stdio(0);cin.tie(0)
#define maxn 100001
const int inf=0x7fffffff;   //无限大
const int MAXN = ;
bool flag[MAXN];
int primes[MAXN], pi;
struct point
{
    int x;
    int y;
};
void GetPrime_1()
{
    int i, j;
    pi = ;
    memset(flag, false, sizeof(flag));
    for (i = ; i < MAXN; i++)
        if (!flag[i])
        {
            primes[i] = ;//素数标识为1
            for (j = i; j < MAXN; j += i)
                flag[j] = true;
        }
}
int vis[maxn];
int main()
{
    GetPrime_1();
    int t;
    cin>>t;
    while(t--)
    {
        memset(vis,,sizeof(vis));
        int n,m;
        cin>>n>>m;
        vis[n]=;
        queue<point> q;
        q.push((point){n,});
        int flag1=;
        while(!q.empty())
        {
            point now=q.front();
            if(now.x==m)
            {
                flag1=now.y;
                break;
            }
            point next;
            for(int i=;i<=;i++)
            {
                next.x=now.x/;
                next.x*=;
                next.x+=i;
                next.y=now.y+;
                if(next.x<||next.x>=)
                    continue;
                if(vis[next.x]==)
                    continue;
                if(next.x==m)
                {
                    flag1=next.y;
                    break;
                }
                if(primes[next.x]==)
                {
                    //cout<<next.x<<endl;
                    vis[next.x]=;
                    q.push(next);
                }

            }
            for(int i=;i<=;i++)
            {
                int temp=now.x%;
                next.x=now.x/;
                next.x*=;
                next.x+=i*;
                next.x+=temp;
                if(next.x<||next.x>=)
                    continue;
                if(vis[next.x]==)
                    continue;
                if(next.x==m)
                {
                    flag1=next.y;
                    break;
                }
                if(primes[next.x]==)
                {
                    //cout<<next.x<<endl;
                    vis[next.x]=;
                    q.push((point){next.x,now.y+});
                }
            }
            for(int i=;i<=;i++)
            {
                int temp=now.x%;
                next.x=now.x/;
                next.x*=;
                next.x+=i*;
                next.x+=temp;
                if(next.x<||next.x>=)
                    continue;
                if(vis[next.x]==)
                    continue;
                if(next.x==m)
                {
                    flag1=next.y;
                    break;
                }
                if(primes[next.x]==)
                {
                    //cout<<next.x<<endl;
                    vis[next.x]=;
                    q.push((point){next.x,now.y+});
                }
            }
            for(int i=;i<=;i++)
            {
                int temp=now.x%;
                next.x=now.x/;
                next.x*=;
                next.x+=i*;
                next.x+=temp;
                if(next.x<||next.x>=)
                    continue;
                if(vis[next.x]==)
                    continue;
                if(next.x==m)
                {
                    flag1=next.y;
                    break;
                }
                if(primes[next.x]==)
                {
                //    cout<<next.x<<endl;
                    vis[next.x]=;
                    q.push((point){next.x,now.y+});
                }
            }
            if(flag1>)
            break;
            q.pop();
        }
    printf("%d\n",flag1);
    }
    return ;
}