POJ-3126-Prime Path

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

每个数字共有四位，每位数字有10种可能的改变值（从0到9），但是最高位不能变为0！将问题转化为图：初始素数和所有经过一位数值改变而得到的新的素数为节点，若素数primeA经过改变后变为新的素数primeB，则A指向B！若目标素数over在图中，则开始素数至目标素数的路径上的边数ans*1即为花费的数目，否则无解！这样一来，问题就可以转化为求从素数start到素数over的最短路径了，使用BFS应该是最为合适的了！

在BFS搜索时应该设立一个数组来存目前得到的所有素数的最短路长度！！！

注意：由于题目要求多组数据输入，所以每次输入完毕后应该将队列清空，再进行广度优先搜索！！！

代码如下：

#include<cstdio>#include<cstring>#include<queue>#include<iostream>using namespace std;const int maxnum=52013;//定义常变量maxnum！int su[maxnum],ss[maxnum],ans;int begin,over;//定义输入端！struct node{    int k,len;//当前值和长度！};queue<node>Q;//建立队列！void prime()//函数，筛选素数法！{    memset(su,0,sizeof(su));    su[0]=su[1]=1;    for(int i=2; i<maxnum; i++)        if(!su[i])            for(int j=2; j*i<maxnum; j++)                su[i*j]=1;}int change(int num,int i,int k){    if(i==1)        return (num/10)*10+k;//钱加尾！    if(i==2)        return num%10+(num/100)*100+k*10;//尾加前加中！    if(i==3)        return num%100+(num/1000)*1000+k*100;    if(i==4)        return num%1000+k*1000;}void path(){    node now,wow,lin;    lin.k=begin;    lin.len=0;    Q.push(lin);    while(!Q.empty())    {        now=Q.front();        Q.pop();        if(now.k==over)//找到        {            ans=now.len;            return ;        }        int knum,klen;        for(int i=1; i<=4; i++)            for(int j=0; j<10; j++)                if(!(i==4&&j==0))                {                    //防止四位数的最高位被改成 0 ！                    knum=change(now.k,i,j);//改变后的值！                    if(su[knum])                        continue;//如果该数不是素数，直接排除掉！                    klen=now.len+1;                    if(klen>=ss[knum])                        continue;//若此时路径长度不是最短的话，直接排除！                    if(knum==over)                    {                        ans=klen;                        return ;                    }                    ss[knum]=klen;                    wow.k=knum;                    wow.len=klen;                    Q.push(wow);                }    }}int main(){    prime();    //cout<<"测试ing……"<<endl;    int T;    cin>>T;    while(T--)    {        while(!Q.empty())            Q.pop();//清空队列！        cin>>begin>>over;        memset(ss,520,sizeof(ss));        path();        if(ans>=0)            cout<<ans<<endl;        else            cout<<"Impossible"<<endl;    }    return 0;}