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

Sample Output

6
7
0

思路：BFS，最先找到的必定是最小解。

 1 #include<iostream>
 2 #include<cstdio>
 3 #include<cstring>
 4 #include<cstdlib>
 5 #include<queue>
 6 #include<cmath>
 7 #define MAX 11111
 8 using namespace std;
 9 int isprime[MAX], pre[MAX];
10 queue<int>q;
11 void Chose_Prime(){
12     isprime[0] = isprime[1] = 0;
13     for(int i = 2;i < MAX;i ++){
14         if(!isprime[i]){
15             for(int j = i + i;j < MAX;j += i)
16                 isprime[j] = 1;
17         }
18     }
19 }
20 int Switch(int num, int i, int j){
21     int target = num/(int)pow(10., 3-i);
22     int temp = target%10;
23     return num += (j - temp)*(int)pow(10., 3-i);
24 }
25 int bfs(int st, int end){
26     int rr;
27     while(!q.empty()) q.pop();
28     q.push(st);
29     pre[st] = 0;
30     while(!q.empty()){
31         int p = q.front();
32         q.pop();
33         for(int i = 0;i < 4;i ++){
34             for(int j = 0;j <= 9;j ++){
35                 if(i + j){
36                     rr = Switch(p, i, j);
37                     if(!isprime[rr] && pre[rr] == -1){
38                         pre[rr] = p;
39                         q.push(rr);
40                         if(rr == end) return true;
41                     }
42                 }
43             }
44         }
45     }
46     return false;
47 }
48 int main(){
49     int T, a, b, ans;
50     Chose_Prime();
51     //freopen("in.c", "r", stdin);
52     cin >> T;
53     while(T--){
54         memset(pre, -1, sizeof(pre));
55         ans = 0;
56         cin >> a >> b;
57         if(a == b){
58             cout << 0 << endl;
59             continue;
60         }
61         if(bfs(a, b)){
62             while(pre[b] != 0){
63                 ans ++;
64                 b = pre[b];
65             }
66             cout << ans << endl;
67         }else{
68             cout << "impossible" << endl;
69         }
70     }
71     return 0;
72 }