1646. Prime Path

单点时限: 2.0 sec

内存限制: 256 MB

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.

输入格式
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).

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

样例
input
3
1033 8179
1373 8017
1033 1033
output
6
7
0
题意：就是每次仅改动上一轮数中的一位数之后，依旧为素数，最多改成到目标素数的次数。

/*
bfs将一轮可改为素数加进去，到了目标素数输出。
*/
#include<bits/stdc++.h>
#define MAX 10000
using namespace std;
typedef pair<string,int> P;
string n,m;
int t,sign[MAX],pri[MAX];//sign标记数是否被用过
void bfs() {queue<P>pq;string x;int step;pq.push({n,0});while(!pq.empty()) {P p=pq.front();pq.pop();x=p.first;step=p.second;if(x==m) {printf("%d\n",step);return ;}for(int i=0; i<=3; i++)for(int j=0; j<=9; j++)if(!(i==0&&j==0)) { //千位数字不能为0string str=x;str[i]='0'+j;if(str!=x&&!sign[atoi(str.c_str())]&&!pri[atoi(str.c_str())]) { //若变换后数字不想等且未使用过且为素数sign[atoi(str.c_str())]++;pq.push({str,step+1});}}}puts("Impossible");
}
int main() {for(int i=2; i<MAX; i++)for(int j=i; i*j<MAX; j++)pri[i*j]=1;cin>>t;while(t--) {fill(sign,sign+MAX,0);cin>>n>>m;sign[atoi(n.c_str())]++;bfs();}
}

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