POJ 3126 Prime Path（BFS）

Prime Path

Time Limit: 1000MS Memory Limit: 65536KTotal Submissions: 12616 Accepted: 7145

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

题目大意：

这道题目是说，给你一个l和r，并且l和r都是素数，问你如何通过最小的步数使得l变成r，并且每次变化只能变一个位置的数字，且每次变化完后都是素数

解题思路：

直接bfs就好，每次将个位，十位，百位，千位的变化后的数字压入队列，然后判断是不是使我们所要求的那个数字，然后记录步数就好了。

代码：

# include<cstdio># include<iostream># include<cstring># include<queue>using namespace std;# define MAX 10000int prime[MAX];int book[MAX];int path[MAX];int l,r;void init(){    for ( int i = 2;i < 10000;i++ )    {        prime[i] = 1;    }    for ( int i = 2;i < 100;i++ )    {        if ( prime[i] )        {            for ( int j = 1;i*j < 10000;j++ )            {                prime[i*j] = 0;            }        }    }}void bfs(){        queue<int>q;        q.push(l);        book[l] = 1;        while ( !q.empty() )        {            int temp = q.front();            q.pop();            if ( temp == r )                break;            for ( int t = 1;t <= 1000;t*=10 )            {                int d = temp;                d/=t;                d%=10;                int rec = temp-d*t;                for ( int j = 0;j <= 9;j++ )                {                    if ( t == 1000&&j==0 )                    {                        continue;                    }                    if ( j!=d )                    {                        int cur = j*t+rec;                        if ( prime[cur]==1&&book[cur]==0 )                        {                            q.push(cur);                            path[cur] = path[temp]+1;                            book[cur] = 1;                        }                    }                }            }        }}int main(void){    int t;cin>>t;    while ( t-- )    {        memset(book,0,sizeof(book));        memset(path,0,sizeof(path));        cin>>l>>r;        init();        bfs();        cout<<path[r]<<endl;    }    return 0;}