POJ ---3126 Prime Path

Prime Path

Time Limit: 1000MS		Memory Limit: 65536K
Total Submissions: 10370		Accepted: 5922

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

3
1033 8179
1373 8017
1033 1033

Sample Output

6
7
0

思路：BFS，最先找到的必定是最小解。

#include<iostream>
#include<cstdio>
#include<cstring>
#include<cstdlib>
#include<queue>
#include<cmath>
#define MAX 11111
using namespace std;
int isprime[MAX], pre[MAX];
queue<int>q;
void Chose_Prime(){
    isprime[0] = isprime[1] = 0;
    for(int i = 2;i < MAX;i ++){
        if(!isprime[i]){
            for(int j = i + i;j < MAX;j += i)
                isprime[j] = 1;
        }
    }
}
int Switch(int num, int i, int j){
    int target = num/(int)pow(10., 3-i);
    int temp = target%10;
    return num += (j - temp)*(int)pow(10., 3-i);
}
int bfs(int st, int end){
    int rr;
    while(!q.empty()) q.pop();
    q.push(st);
    pre[st] = 0;
    while(!q.empty()){
        int p = q.front();
        q.pop();
        for(int i = 0;i < 4;i ++){
            for(int j = 0;j <= 9;j ++){
                if(i + j){
                    rr = Switch(p, i, j);
                    if(!isprime[rr] && pre[rr] == -1){
                        pre[rr] = p;
                        q.push(rr);
                        if(rr == end) return true;
                    }
                }
            }
        }
    }
    return false;
}
int main(){
    int T, a, b, ans;
    Chose_Prime();
    //freopen("in.c", "r", stdin);
    cin >> T;
    while(T--){
        memset(pre, -1, sizeof(pre));
        ans = 0;
        cin >> a >> b;
        if(a == b){
            cout << 0 << endl;
            continue;
        }
        if(bfs(a, b)){
            while(pre[b] != 0){
                ans ++;
                b = pre[b];
            }
            cout << ans << endl;
        }else{
            cout << "impossible" << endl;
        }
    }
    return 0;
}