I - 09（广搜）

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

题意：

给你两个四位数的素数，问第一个经过几次变换能变换成第二个，并且变换变换过程中保证变换的数仍为素数！！

思路：

首先，筛四位数的素数，然后，利用创建一个队列，和数组，用来存放数字，如果出现过，将数组对应项置为1，然后进行广搜，符合条件的入队！！！！

代码：

#include <iostream>
#include <cstdio>
#include <cstring>
#include <algorithm>
#include <cmath>
#include <string>
#include <map>
#include <stack>
#include <vector>
#include <set>
#include <queue>
#define maxn 15
#define MAXN 10005
#define mod 1000000007
#define INF 0x3f3f3f3f
#define exp 1e-6
#define pi acos(-1.0)
using namespace std;
int isprime[MAXN];  //素数数组
int v[MAXN];
int flag=0;
struct node   //结构体，存放调整的次数，数字，以及每位数
{
    int w[5];
    int s;
    int num;
};
void go()
{
    int i,j;
    for(i=0;i<MAXN;i++)
        isprime[i]=1;
    isprime[0]=isprime[1]=0;
    for (i=2;i<MAXN;i++)
    {
        if(isprime[i]==1)
        {
        for(int j=i+i;j<MAXN;j+=i)
        {
           isprime[j]=0;
        }
        }
    }
}
int getdata(int a[],int b,int c)  //生成新得到数
{
    int ans=0;
    for(int i=0;i<4;i++)
    {
        if(i==b)
            ans=ans*10+c;
        else
            ans=ans*10+a[i];
    }
    return ans;
}
void work(int first,int second)  //搜索
{
    queue<node>ss;
    int i,j,k,num1;
    node start,endd;
    memset(v,0,sizeof(v));
    start.num=first;
    start.s=0;
    start.w[3]=first%10;
    start.w[2]=(first/10)%10;
    start.w[1]=(first/100)%10;
    start.w[0]=(first/1000)%10;
    ss.push(start);
    while(!ss.empty())
    {
        endd=ss.front();
        ss.pop();
        if(endd.num==second)
        {
            cout<<endd.s<<endl;
            flag=1;                     //定个flag，避免找不到的情形
            break;
        }
        for(i=0;i<4;i++)
        {
            for(j=0;j<=9;j++)
            {
                if(i==0&&j==0)
                    continue;
                if(endd.w[i]==j)
                    continue;
                    else
                    {
                        num1=getdata(endd.w,i,j);

                        if(isprime[num1]&&!v[num1])  //同时满足
                        {
                            v[num1]=1;
                            for(k=0;k<4;k++)
                            {
                                start.w[k]=endd.w[k];
                            }
                            start.num=num1;
                            start.w[i]=j;
                            start.s=endd.s+1;
                            ss.push(start);  //入队
                        }
                     }
            }
        }
    }

}
int main()
{
    ios::sync_with_stdio(false);
    int first,second;
    int t;
    go();
    cin>>t;
    while(t--)
    {
        flag=0;
        cin>>first>>second;
        work(first,second);
        if(flag==0)
        {
            cout<<"Impossible"<<endl;
        }
    }
    return 0;
}

多做题！！！！！！！！！！