描述

传送门：poj-3126
传送门：swustoj-2

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.

Examples

intput
1
2
3
4
3
1033 8179
1373 8017
1033 1033
output
1
2
3
6
7
0

题目大意

如果两个四位的素数有且只有一位数字不一样时，这两两个素数可以互相变化成对方。

每组数据给你两个素数，求从一个到另一个素数需要变化的最小次数。

思路

用素数筛预处理出一万内所有素数。

$O(N^2)$预处理出边，建图完毕。

链式向前星存图。

最后从起点到终点跑一遍spfa，求得最短路，即答案。

用bfs跑也是可以过的。

代码

#include<iostream>
#include<vector>
#include<queue>
#include<string.h>
using namespace std;
#define rep(i,a,n) for(int i=a;i<n;i++)
#define repd(i,a,n) for(int i=n-1;i>=a;i--)
#define CRL(a) memset(a,0,sizeof(a))
#define CRL2(a,x) memset(a,x,sizeof(a))
const int mod = 1e9 + 7;
const int N = 1e4+5;

int prime[N] = { 0 }, num_prime = 0;
int isNotPrime[N] = { 1, 1 }, BeginP=-1;
struct node {int To, S, Next;}; //链式向前星存图

vector <node> Edge;
int Begin[N], Inque[N], Dis[N], n;

void GetPrim() {        //线筛
    rep(i,2,N) {
        if (!isNotPrime[i]) {
            prime[num_prime++] = i;
            if (i>1000&&BeginP==-1)
                BeginP = num_prime-1;       //BeginP用来存第一个四位数的，注意是-1这个点wa了好久
        }

        for (int j = 0; j<num_prime&&i*prime[j]<N; j++) {
            isNotPrime[i*prime[j]] = 1;
            if (!(i%prime[j]))
                break;
        }
    }
}

void add(int a, int b) {        //加边
    node tem;
    tem.To = b;
    tem.S = 1;
    tem.Next = Begin[a];
    Begin[a] = Edge.size();
    Edge.push_back(tem);
}

bool Judge(int x, int y) {  //判断是否只有一位不同
    int Flag = 0;
    rep(i, 0, 4) {
        if (x % 10 != y % 10) Flag++;
        x /= 10; y /= 10;
    }
    return Flag==1;
}

void spfa(int s) {//最短路
    rep(i,0,N+1) Dis[i] = 9999999;Dis[s] = 0;
    CRL(Inque);Inque[s] = 1;
    queue<int> Q;Q.push(s);
    while (!Q.empty()) {
        int T = Q.front();
        Q.pop();Inque[T] = 0;
        for (int i = Begin[T]; i != -1; i = Edge[i].Next) {
            if (Dis[Edge[i].To]>Dis[T] + Edge[i].S) {
                Dis[Edge[i].To] = Dis[T] + Edge[i].S;
                Q.push(Edge[i].To);
                Inque[Edge[i].To] = 1;
            }
        }
    }
}

int main() {
    ios::sync_with_stdio(false);
    int x, Star, End, T;
    GetPrim();

    CRL2(Begin, -1);
    rep(i,BeginP,num_prime)
        rep(j,i,num_prime)
            if (Judge(prime[i], prime[j])) {
                add(prime[i], prime[j]);
                add(prime[j], prime[i]);
            }
    cin >> T;
    while (T--) {
        cin >> Star >> End;
        spfa(Star);
        if(Dis[End]!=9999999)
            cout << Dis[End] << endl;
        else
            cout << "Impossible" << endl;
    }
    return 0;
}