E - Prime Path （bfs+找素数）

E - Prime Path

 

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

此题是将前一个数通过改变每一个位上的数来变成最后的数，而且改变的数也要是素数

//Full of love and hope for life


#include <iostream>
#include <string.h>
#include <math.h>
#include <algorithm>
#include <stdio.h>
#include <queue>
using namespace std;

int prime[10010];

void init()//将所有素数找出来，并且标记
{
    prime[1]=1;
    for(int i=2; i<=100; i++){
        if(prime[i]==0){
            for(int j=i; j<=10000; j+=i){
                prime[j]=1;
            }
        }
    }
}

struct node
{
    int x,step;
};

node n,m;
int b,c;
int flag[10010];

void bfs()
{
    queue<node>q;
    n.x=b;
    n.step=0;
    flag[n.x]=1;
    q.push(n);
    while(!q.empty()){
        m=q.front();
        q.pop();
        if(m.x==c){
            cout << m.step << endl;
            return ;
        }
        for(int i=1; i<=4; i++){
            for(int j=0; j<=9; j++){
                if(i==1){
                    n.x=m.x-m.x%10+j;//个位
                }
                if(i==2){
                    n.x=m.x+(j-m.x%100/10)*10;//十位
                }
                if(i==3){
                    n.x=m.x+(j-m.x%1000/100)*100;//百位
                }
                if(i==4&&j!=0){
                    n.x=m.x+(j-m.x/1000)*1000;//千位，并且不能为0
                }
                if(prime[n.x]==0&&flag[n.x]==0){
                    flag[n.x]=1;
                    n.step=m.step+1;
                    q.push(n);
                }
            }
        }
    }
    cout << "Impossible" << endl;
    return ;
}

int main()
{
    init();
    int a;
    cin >> a;
    while(a--){
        memset(flag,0,sizeof(flag));
        cin >> b >> c;
        if(b==c){
            cout << 0 << endl;
        }
        else{
            bfs();
        }
    }
    return 0;
}