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http://acm.hdu.edu.cn/showproblem.php?pid=1973
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.
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).
One line for each case, either with a number stating the minimal cost or containing the word Impossible.
3
1033 8179
1373 8017
1033 1033
6
7
0
1 #include<algorithm> 2 #include<iostream> 3 #include<cstdlib> 4 #include<cstring> 5 #include<cstdio> 6 #include<vector> 7 #include<queue> 8 #include<map> 9 using std::cin; 10 using std::cout; 11 using std::endl; 12 using std::find; 13 using std::sort; 14 using std::map; 15 using std::pair; 16 using std::queue; 17 using std::vector; 18 using std::reverse; 19 #define pb(e) push_back(e) 20 #define sz(c) (int)(c).size() 21 #define mp(a, b) make_pair(a, b) 22 #define all(c) (c).begin(), (c).end() 23 #define iter(c) decltype((c).begin()) 24 #define cls(arr,val) memset(arr,val,sizeof(arr)) 25 #define cpresent(c, e) (find(all(c), (e)) != (c).end()) 26 #define rep(i, n) for (int i = 0; i < (int)(n); i++) 27 #define tr(c, i) for (iter(c) i = (c).begin(); i != (c).end(); ++i) 28 const int Max_N = 10010; 29 typedef unsigned long long ull; 30 int start, end; 31 namespace work { 32 struct Node { 33 int v, s; 34 Node(int i = 0, int j = 0) :v(i), s(j) {} 35 }; 36 bool vis[Max_N], Prime[Max_N]; 37 inline bool isPrime(int n) { 38 for (int i = 2; (ull)i * i <= n; i++) { 39 if (0 == n % i) return false; 40 } 41 return n != 1; 42 } 43 inline void init() { 44 for (int i = 1; i < Max_N; i++) { 45 Prime[i] = isPrime(i); 46 } 47 } 48 inline void bfs() { 49 char buf[10], str[10]; 50 cls(vis, false); 51 queue<Node> que; 52 que.push(Node(start, 0)); 53 vis[start] = true; 54 while (!que.empty()) { 55 Node tmp = que.front(); que.pop(); 56 if (tmp.v == end) { printf("%d\n", tmp.s); return; } 57 sprintf(buf, "%d", tmp.v); 58 reverse(buf, buf + 4); 59 for (int i = 0; buf[i] != ‘\0‘; i++) { 60 rep(j, 10) { 61 strcpy(str, buf); 62 str[i] = j + ‘0‘; 63 reverse(str, str + 4); 64 int v = atoi(str); 65 if (vis[v] || !Prime[v]) continue; 66 que.push(Node(v, tmp.s + 1)); 67 vis[v] = true; 68 } 69 } 70 } 71 puts("Impossible"); 72 } 73 } 74 int main() { 75 #ifdef LOCAL 76 freopen("in.txt", "r", stdin); 77 freopen("out.txt", "w+", stdout); 78 #endif 79 int t; 80 work::init(); 81 scanf("%d", &t); 82 while (t--) { 83 scanf("%d %d", &start, &end); 84 work::bfs(); 85 } 86 return 0; 87 }
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原文地址：http://www.cnblogs.com/GadyPu/p/4608375.html
