F - Prime Path

题目描述

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

#include<iostream>
#include<algorithm>
#include<stdio.h>
#include<cstring>
#include<queue>
#define maxn 10000
#define INF 0x3f3f3f3f
using namespace std;
typedef long long ll;
int n,m;
bool prime[maxn +5];
int dis[maxn + 5];
struct p{int x,step;
}cur, nxt;
void dfs()
{queue<p> q;cur.x = n;cur.step = 0;q.push(cur);while(q.size()){cur = q.front();q.pop();if(cur.x == m){cout << cur.step <<endl;return ;}nxt.step = cur.step + 1;for(int i = 0; i < 4; i++){for(int j = 0; j <= 9; j++){if(i == 0)nxt.x = cur.x - cur.x % 10 + j;if(i == 1)nxt.x = cur.x + (j - cur.x%100/10)*10;if(i == 2)nxt.x = cur.x + (j - cur.x/100 % 10)*100;if(i == 3 && j != 0)nxt.x = cur.x + (j - cur.x/1000) *1000;if(prime[nxt.x] && !dis[nxt.x]){dis[nxt.x] = 1;q.push(nxt);}}}}cout << "Impossible" <<endl;return ;
}
int main()
{int t;cin >> t;memset(prime, true, sizeof prime);for(int i = 2; i <=100; i++){for(int j = 1; j *i <= 10000; j++){prime[i * j] = false;}}prime[1] = false;while(t--){memset(dis, 0, sizeof dis);cin >> n >> m;if(n == m)cout << '0' << endl;else {dis[n] = 1;dfs();}}return 0;
}

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