POJ-3126-Prime Path（BFS）

Prime Path

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：
在1000到10000的範圍內，輸入兩個數，x,y。 y>x,現在要讓你把x變成y。改變x四位數中的任何一位之後如果它爲素數，就以他爲基準繼續變，直到它變成y，也就是以素數爲臺階。
這個題感覺還好，一看過去就知道這是用bfs（），因爲說的是最短麼，所以就用 bfs() 就好了，主要就是這些操作，一個數我們要四次操作。每次操作10個方向，還是很麻煩的。（有的大佬是打表，但是我覺得不打表也成。）
完整代碼：

#include<iostream>
#include<cstring>
#include<queue>
#include<cmath>
using namespace std;
int n,x,y;
int Prime[10005];
int t[10005];
void init() //篩選素數，將素數標記
{
    Prime[1]=1;
    for(int i=2;i<=sqrt(10000);i++)
    {
        if(Prime[i]==0)
        {
            for(int j=i+i;j<=10000;j+=i)
            {
                Prime[j]=1;
            }
        }
    }
}
struct node
{
    int w,s;    //位，步數
};
void bfs()  //搜索
{
    node Now,Next;
    queue<node>q;
    Now.w=x;
    Now.s=0;
    q.push(Now);
    while(!q.empty())
    {
        Now=q.front();
        q.pop();
        Next.s=Now.s+1;
        if(Now.w==y)
        {
            cout<<Now.s<<endl;
            return ;
        }
        for(int i=0;i<4;i++)
        {
            for(int j=0;j<=9;j++)
            {
                if(i==0)
                {
                    Next.w=Now.w-Now.w%10+j;
                }
                else if(i==1)
                {
                    Next.w=Now.w+(j-Now.w%100/10)*10;
                }
                else if(i==2)
                {
                    Next.w=Now.w+(j-Now.w%1000/100)*100;
                }
                else if(i==3&&j!=0)
                {
                    Next.w=Now.w+(j-Now.w/1000)*1000;
                }
                if(Prime[Next.w]==0&&t[Next.w]==0)
                {
                    q.push(Next);
                    t[Next.w]=1;
                }
            }
        }
    }
    cout<<"Impossible"<<endl;
    return ;
}
int main()
{
    init();
    cin>>n;
    while(n--)
    {
        memset(t,0,sizeof(t));
        cin>>x>>y;
        if(x==y)
        {
            cout<<"0"<<endl;
            continue;
        }
        t[x]=1;
        bfs();
    }
    return 0;
}

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原題鏈接：
https://vjudge.net/problem/POJ-3126
http://poj.org/problem?id=3126