poj 3126.Prime Path（bfs）

题目链接：http://poj.org/problem?id=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.

1033173337333739377987798179

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

题目大意：简单来说，就是有t组数据，然后给你两个四位的素数数 n 和 m 这两个数字不考虑前导0 的情况。然后每次你可以改变某一位上的数字，但是要求改变之后的数字也是素数，问至少几次 可以使n变成m 。 这个思路就有了，可以用bfs的思想。
看下我的代码 和注释，你们肯定会懂得。

#include <stdio.h>#include <string.h>#include <queue>#include <algorithm>using namespace std;int vis[10001];            //数字的访问标记，因为访问过的数字　如果变回去的话　肯定增加了步数int n,m;                //两个素数int res;                //记录最后的结果int pow(int a,int b)            //自己写的指数函数{    int res = 1;    for (int i = 0; i < b ; i++ )    {        res *= a;    }    return res;}struct sta            {    int n;                //记录数字    int count;            //记录变了几次}num;int check(int num)                //判断数字符不符合要求{    int flag = 0;    int temp = (num / 2) + 1;    if ( vis[num] == 1 || num > 9999 || num < 1000)            //如果访问过了 就直接跳出        return 0;     for (int i = 2 ; i <= temp ; i++ )                //求这个数是不是素数    {        if ( num % i == 0 )        {            flag =1;            break;        }    }    if ( flag == 0 && vis[num] == 0 && num >= 1000 && num <= 9999)    {        return 1;    }    else     {        return 0;    }}void  bfs(sta s){    queue <sta> q;    s.count = 0;    vis[s.n] = 1;    q.push(s);            //刚开始的数字入队    sta now,next;    while (!q.empty())        {        now = q.front();        q.pop();            if (now.n == m )            //如果已经变化到结果,则退出        {            res = now.count;            return ;        }        for (int i = 1 ; i <= 4 ; i++ )        {            for (int j = 0 ; j <= 9 ; j++ )            {                next.n = now.n -( (now.n/pow(10,i-1)) % 10 ) * pow(10,i-1) + pow(10,i-1)* j;  //将某一位的数字先变成0  之后再变成1，2，3，。。。相当于直接改变                next.count = now.count + 1;                //步数增加                if (check(next.n))                //如果符合要求，入队                {                    vis[next.n] = 1;                    q.push(next);                }            }        }    }    return ;}int main(){    int t;    scanf("%d",&t);    while ( t-- )    {        res = 0;        memset(vis,0,sizeof(vis));            //每次输入都要初始化vis        scanf("%d%d",&n,&m);        num.n = n;        num.count = 0;        bfs(num);        printf("%d\n",res);                //输出结果    }    return 0;}