Swust OJ 0002 Prime Path

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Swust OJ 解题报告

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首相因更换办公室门上的四位数素数号码感到困扰，要求通过改变单个数字保持素数状态且成本最低。财政大臣介入，提出每个数字成本为一英镑，并引入编程竞赛来解决问题。题目描述给出，测试用例包含两组四位数素数，需要找到最小花费的素数路径，且不能重复使用已覆盖的数字。

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.

Description

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).

Input

One line for each case, either with a number stating the minimal cost or containing the word Impossible.

Output

Sample Input

6
7
0

/*
 * 题意：给2个4位素数a,b 
    1. 改变a的一位  得到c 并且保证c也是素数
	2. 改变的c的一位 得到d 并且保证d也是素数..重复以上操作 问最后得到b最短需要几步。

	bfs即可。。并不想多说
	吐槽一下：神奇的Swust OJ  测试数据弱的要死- - 我忘了输出不能到达b的时候 输出Impossible 居然AC了。。我还能说什么。。
*/
#include <iostream>
#include <cstdio>
#include <cstdlib>
#include <cstring>
#include <string>
#include <algorithm>
#include <stack>
#include <cmath>
#include <queue>
using namespace std;
int a, b;
bool vis[10001];
struct Node {
	int val;
	int step;
};
bool isPrime(int n){
	for (int i = 3; i <= sqrt(n); i++) {
		if (n%i == 0)
			return 0;
	}
	return 1;
}
void bfs() {
	memset(vis, 0, sizeof(vis));
	queue<Node>q;
	Node l;
	l.val = a;
	l.step = 0;
	q.push(l);
	vis[a] = 1;
	while (!q.empty()) {
		Node p = q.front();
		q.pop();
		if (p.val == b) {
			cout << p.step << endl;
			break;
		}
		Node l;
		for (int i = 1; i <= 9; i++)
		{
			int c = i * 1000 + (p.val % 1000);
			if (!vis[c] && isPrime(c)) {
				l.val = c;
				l.step = p.step + 1;
				vis[c] = 1;
				q.push(l);
			}
		}
		for (int i = 0; i <= 9; i++)
		{
			int c = i * 100 + p.val % 100 + (p.val/1000)*1000;
			if (!vis[c] && isPrime(c)) {
				l.val = c;
				l.step = p.step + 1;
				vis[c] = 1;
				q.push(l);
			}
		}
		for (int i = 0; i <= 9; i++)
		{
			int c = i * 10 + p.val % 10 + (p.val / 100) * 100;
			if (!vis[c] && isPrime(c)) {
				l.val = c;
				l.step = p.step + 1;
				vis[c] = 1;
				q.push(l);
			}
		}
		for (int i = 1; i <= 9; i+=2)
		{
			int c = i + (p.val / 10) * 10;
			if (!vis[c] && isPrime(c)) {
				l.val = c;
				l.step = p.step + 1;
				vis[c] = 1;
				q.push(l);
			}
		}

	}
	if (q.empty()) {
		cout << "Impossible" << endl;
	}
}
int main(){
	int t;
	cin >> t;
	while (t--) {
		cin >> a >> b;
		bfs();
	}
	return 0;
}