POJ-3126
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.
— 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.
1033The 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.
1733
3733
3739
3779
8779
8179
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
AC代码为：
1 #include <iostream> 2 #include <cstdio> 3 #include <cstring> 4 #include <algorithm> 5 #include <cmath> 6 #include <string> 7 #include <queue> 8 using namespace std; 9 #define maxn 10005 10 bool visit[maxn]; 11 int m, n, a, b, c, d; 12 struct Node 13 { 14 int p[4]; int step; 15 }; 16 bool is_prime(int x) 17 { 18 int sum = 0; 19 if (x == 1) return false; 20 if (x == 2) return true; 21 for (int i = 2; i <= sqrt(x); i++) 22 { 23 if (x%i == 0) sum++; 24 } 25 if (sum) return false; else return true; } 26 int BFS() 27 { 28 Node st, now; memset(visit, false, sizeof(visit)); 29 queue<Node>Q; 30 while (!Q.empty()) Q.pop(); 31 visit[m] = true; st.p[0] = m / 1000; st.p[1] = (m / 100) % 10; st.p[2] = (m / 10) % 10; st.p[3] = m % 10; st.step = 0; Q.push(st); 32 while (!Q.empty()) 33 { 34 st = Q.front(); Q.pop(); 35 if (st.p[0] == a && st.p[1] == b && st.p[2] == c && st.p[3] == d) { return st.step; } 36 for (int i = 0; i <= 3; i++) 37 { 38 for (int j = 0; j<10; j++) 39 { 40 if (st.p[i] == j) continue; 41 if (i == 0 && j == 0) continue; now.p[0] = st.p[0]; now.p[1] = st.p[1]; now.p[2] = st.p[2]; now.p[3] = st.p[3]; now.p[i] = j; int x = now.p[0] * 1000 + now.p[1] * 100 + now.p[2] * 10 + now.p[3]; if (is_prime(x) && !visit[x]) { visit[x] = true; now.step = st.step + 1; Q.push(now); } } } } 42 return -1; 43 } 44 int main() 45 { 46 int N; scanf("%d", &N); 47 while (N--) 48 { 49 scanf("%d%d", &m, &n); a = n / 1000; b = (n / 100) % 10; c = (n / 10) % 10; d = n % 10; int ans = BFS(); 50 if (ans == -1) printf("Impossible\n"); else printf("%d\n", ans); } 51 return 0; 52 }
