hdu 1973 Prime Path

题目连接

http://acm.hdu.edu.cn/showproblem.php?pid=1973

Prime Path

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

3
1033 8179
1373 8017
1033 1033

Sample Output

6
7
0

bfs。。。

 #include<algorithm>
 #include<iostream>
 #include<cstdlib>
 #include<cstring>
 #include<cstdio>
 #include<vector>
 #include<queue>
 #include<map>
 using std::cin;
 using std::cout;
 using std::endl;
 using std::find;
 using std::sort;
 using std::map;
 using std::pair;
 using std::queue;
 using std::vector;
 using std::reverse;
 #define pb(e) push_back(e)
 #define sz(c) (int)(c).size()
 #define mp(a, b) make_pair(a, b)
 #define all(c) (c).begin(), (c).end()
 #define iter(c) decltype((c).begin())
 #define cls(arr,val) memset(arr,val,sizeof(arr))
 #define cpresent(c, e) (find(all(c), (e)) != (c).end())
 #define rep(i, n) for (int i = 0; i < (int)(n); i++)
 #define tr(c, i) for (iter(c) i = (c).begin(); i != (c).end(); ++i)
 const int Max_N = ;
 typedef unsigned long long ull;
 int start, end;
 namespace work {
     struct Node {
         int v, s;
         Node(int i = , int j = ) :v(i), s(j) {}
     };
     bool vis[Max_N], Prime[Max_N];
     inline bool isPrime(int n) {
         for (int i = ; (ull)i * i <= n; i++) {
             if ( == n % i) return false;
         }
         return n != ;
     }
     inline void init() {
         for (int i = ; i < Max_N; i++) {
             Prime[i] = isPrime(i);
         }
     }
     inline void bfs() {
         char buf[], str[];
         cls(vis, false);
         queue<Node> que;
         que.push(Node(start, ));
         vis[start] = true;
         while (!que.empty()) {
             Node tmp = que.front(); que.pop();
             if (tmp.v == end) { printf("%d\n", tmp.s); return; }
             sprintf(buf, "%d", tmp.v);
             reverse(buf, buf + );
             for (int i = ; buf[i] != '\0'; i++) {
                 rep(j, ) {
                     strcpy(str, buf);
                     str[i] = j + '';
                     reverse(str, str + );
                     int v = atoi(str);
                     if (vis[v] || !Prime[v]) continue;
                     que.push(Node(v, tmp.s + ));
                     vis[v] = true;
                 }
             }
         }
         puts("Impossible");
     }
 }
 int main() {
 #ifdef LOCAL
     freopen("in.txt", "r", stdin);
     freopen("out.txt", "w+", stdout);
 #endif
     int t;
     work::init();
     scanf("%d", &t);
     while (t--) {
         scanf("%d %d", &start, &end);
         work::bfs();
     }
     return ;
 }