Poj3126

Prime Path

Time Limit: 1000MS		Memory Limit: 65536K
Total Submissions: 16201		Accepted: 9151

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

Source

Northwestern Europe 2006

大致题意：

给定两个四位素数a  b，要求把a变换到b

变换的过程要保证  每次变换出来的数都是一个 四位素数，而且当前这步的变换所得的素数  与  前一步得到的素数  只能有一个位不同，而且每步得到的素数都不能重复。 

求从a到b最少需要的变换次数。无法变换则输出Impossible

思路：Eratosthenes筛法+bfs

#include<cstdio>
#include<cstring>
#include<iostream>
#include<algorithm>
#include<queue>
using namespace std;
#define N 10010
int prime[N],a[N],vis[N];
int time,tot;
int n=10000,m,p;
struct node{
    int value;
    int deep;
};
bool hd(int n,int m){//判断只差一位 
    int sum=0;
    int temp1,temp2;
    while(n!=0){
        temp1=n%10;
        temp2=m%10;
        if(temp1!=temp2)
            sum++;
        n/=10;
        m/=10;
    }
    if(sum==1)
        return true;
    return false;
}
inline int bfs(){
    memset(vis,0,sizeof vis);
    queue<node>que;
    node now,next;
    now.deep=0;
    vis[now.value=n]=1;
    que.push(now);
    while(!que.empty()){
        now=que.front();
        que.pop();
        if(now.value==m){
            time=now.deep;//记录 
            return 1;
        }
        for(int i=tot;i>=1;i--){
            if(!vis[prime[i]]&&hd(prime[i],now.value)){
                vis[prime[i]]=1;
                next.value=prime[i];
                next.deep=now.deep+1;
                que.push(next); 
            }
        }
    }
    return 0; 
}
int main(){
    int t;//prime[]存1000-9999的素数 
    for(int i=2;i<=n;i++)
      if(!vis[i]){
          a[tot++]=i;
          for(int j=i*2;j<=n;j+=i)
            vis[j]=1;
      }
    for(int i=0;i<tot;i++)
        if(a[i]>1000)  
           prime[++p]=a[i];
    scanf("%d",&t);
    while(t--){
        scanf("%d%d",&n,&m);
        if(bfs())
            printf("%d\n",time);
        else
            printf("Impossible\n");
    }
    return 0;
}