CDZSC_2015寒假新人（4）——搜索 I

Time Limit:1000MS     Memory Limit:65536KB     64bit IO Format:%I64d & %I64u

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


思路：这题的意思也很容易懂，就是给出初始的4位素数后最终的4位素数，通过每次变换个十百千位其中一个（而且变后的数依然是素数）达到最终的4位素数
我认为我的代码比较巧妙的是fun函数。。。。。

#include <iostream>
#include <cstdio>
#include <cstring>
#include <queue>
#include <cmath>
using namespace std;
#define N 10010
bool prime[N];
int vis[N],step[N];
int aa[]={1,10,100,1000};
void table()
{
    int i,j;
    memset(prime,true,sizeof(prime));
    prime[0]=false;
    prime[1]=false;
    for(i=2; i<N; i++)
    {
        if(prime[i])
        {
            for(j=2*i; j<N; j+=i)
            {
                prime[j]=false;
            }
        }
    }
}
int fun(int num,int k)//这个函数是让num的K位变成0，例如num=1033，k=1，那么输出的就是1030
{
    int p[5],q=0,i;
    while(num>0)
    {
        p[q++]=num%10;
        num/=10;
    }
    p[k]=0;
    for(i=0; i<q; i++)
    {
        num+=p[i]*aa[i];
    }
    return num;
}
int bfs(int a,int b)
{
    int c,temp,i,j,num;
    memset(vis,0,sizeof(vis));
    memset(step,0,sizeof(step));
    vis[a]=1;
    queue<int>q;
    q.push(a);
    while(!q.empty())
    {
        c=q.front();
        q.pop();
        /*printf("%d\n",c);*/
        if(c==b)
        {
            break;
        }
        for(i=0; i<4; i++)
        {
            num=fun(c,i);
            for(j=0; j<=9; j++)
            {
                if(i==3&&j==0)
                {
                    continue;
                }
                temp=num+(j*aa[i]);
                if(prime[temp]&&!vis[temp])
                {
                    vis[temp]=1;
                    step[temp]=step[c]+1;
                    /*printf("%d %d\n",temp,step[temp]);*/
                    q.push(temp);
                }
            }
        }
    }
    return step[b];
}
int main()
{
#ifdef CDZSC_OFFLINE
    freopen("in.txt","r",stdin);
#endif
    int t,a,b,sum;
    table();
    scanf("%d",&t);
    while(t--)
    {
        scanf("%d%d",&a,&b);
        if(a==b)
        {
            printf("0\n");
            continue;
        }
        sum=bfs(a,b);
        printf("%d\n",sum);
    }
    return 0;
}