POJ - 3126 Prime Path 素数筛选+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

题意：题目较长，实际就是给你两个四位素数，让你每次只能更改第一个素数的其中一位，更改后要求也是素数且位数不变，问你至少需要更改几次才能变成第二个素数。无解输出Impossible。
思路：本题涉及到素数，每次更改后均需要判断，所以避免重复计算，在程序开始先用筛法把每个四位数的素数性提前存到数组prime。之后分别更改一位数值（第一位不可能是0，最后一位只能是奇数），记录下变更次数即可。

#include<stdio.h>
#include<string.h>
#include<queue>
using namespace std;

int prime[],bo[];
struct Node{
    int x,s;
}node;

int main()
{
    int t,a,b,f,i,j;
    prime[]=;
    for(i=;i<=;i++){
        if(!prime[i]){
            for(j=;i*j<=;j++){
                prime[i*j]=;     //素数筛选
            }
        }
    }
    scanf("%d",&t);
    while(t--){
        queue<Node> q;
        memset(bo,,sizeof(bo));
        scanf("%d%d",&a,&b);
        if(a==b) printf("0\n");
        else{
            bo[a]=;
            node.x=a;
            node.s=;
            q.push(node);
            f=;
            while(q.size()){
                int tx=q.front().x;
                for(i=;i<=;i++){
                    if(i!=tx%&&!prime[tx-tx%+i]&&bo[tx-tx%+i]==){
                        if(tx-tx%+i==b){
                            f=q.front().s+;
                            break;
                        }
                        bo[tx-tx%+i]=;
                        node.x=tx-tx%+i;
                        node.s=q.front().s+;
                        q.push(node);
                    }
                    if(i!=tx/%&&!prime[tx-tx/%*+i*]&&bo[tx-tx/%*+i*]==){
                        if(tx-tx/%*+i*==b){
                            f=q.front().s+;
                            break;
                        }
                        bo[tx-tx/%*+i*]=;
                        node.x=tx-tx/%*+i*;
                        node.s=q.front().s+;
                        q.push(node);
                    }
                    if(i!=tx/%&&!prime[tx-tx/%*+i*]&&bo[tx-tx/%*+i*]==){
                        if(tx-tx/%*+i*==b){
                            f=q.front().s+;
                            break;
                        }
                        bo[tx-tx/%*+i*]=;
                        node.x=tx-tx/%*+i*;
                        node.s=q.front().s+;
                        q.push(node);
                    }
                    if(i!=&&i!=tx/&&!prime[tx-tx/*+i*]&&bo[tx-tx/*+i*]==){
                        if(tx-tx/*+i*==b){
                            f=q.front().s+;
                            break;
                        }
                        bo[tx-tx/*+i*]=;
                        node.x=tx-tx/*+i*;
                        node.s=q.front().s+;
                        q.push(node);
                    }
                }
                if(f!=) break;
                q.pop();
            }
            if(f==) printf("Impossible\n");
            else printf("%d\n",f);
        }
    }
    return ;
}