poj 3126

Prime Path

Time Limit: 1000MS Memory Limit: 65536KTotal Submissions: 11143 Accepted: 6325

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

31033 81791373 80171033 1033

Sample Output

670

Source

Northwestern Europe 2006

AC代码：

#include<iostream>#include<cmath>#include<queue>using namespace std;struct Node{    int m,t;    Node(int M,int T):m(M),t(T){}};int prime[10010];int sign[10010];queue <Node> qu;void isprime(){    for(int i=1001;i<=9999;i++){        int j;        for(j=2;j<=(int)sqrt(i+0.5);j++)            if(i%j==0){                prime[i]=0;                break;            }        if(j>(int)sqrt(i+0.5))            prime[i]=1;    }}int main(){    //freopen("in.txt","r",stdin);    //freopen("out.txt","w",stdout);    isprime();    int T; cin>>T;    while(T--){        int n,k,ans=99999;        cin>>n>>k;        for(int i=1001;i<=9999;i++)            sign[i]=0;        while(!qu.empty())            qu.pop();        qu.push(Node(n,0));        sign[n]=1;        while(!qu.empty()){            Node tmp=qu.front(); qu.pop();            if(tmp.m==k){                ans=min(ans,tmp.t);                //break;            }            int tp,tt;            tp=tmp.m;            tp/=10; tp*=10;            for(int i=0;i<=9;i++){             //变换个位数字                 if(prime[tp] && !sign[tp]){                    sign[tp]=1;                    //cout<<tp<<' ';                    qu.push(Node(tp,tmp.t+1));                }                tp+=1;            }            tp=tmp.m;            tt=tp%10;            tp/=100; tp*=100; tp+=tt;            for(int i=0;i<=9;i++){                       //变换十位数字                 if(prime[tp] && !sign[tp]){                    sign[tp]=1;                    //cout<<tp<<' ';                    qu.push(Node(tp,tmp.t+1));                }                tp+=10;            }            tp=tmp.m;            tt=tp%100;            tp/=1000; tp*=1000; tp+=tt;            for(int i=0;i<=9;i++){                   //变换百位数字                 if(prime[tp] && !sign[tp]){                    sign[tp]=1;                    //cout<<tp<<' ';                    qu.push(Node(tp,tmp.t+1));                }                tp+=100;            }            tp=tmp.m;            tp=tp%1000;            for(int i=1;i<=9;i++){                   //变换千位数字                 tp+=1000;                if(prime[tp] && !sign[tp]){                    sign[tp]=1;                    //cout<<tp<<' ';                    qu.push(Node(tp,tmp.t+1));                }            }        }        //cout<<endl;        if(ans==99999)            cout<<"Impossible"<<endl;        else            cout<<ans<<endl;    }    return 0;}