http://poj.org/problem?id=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.
http://www.hankcs.com/program/cpp/poj-3126-prime-path.html
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.
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 0http://www.2cto.com/kf/201401/272400.html
题意：给定两个素数n和m，要求把n变成m，每次变换时只能变一个数字，即变换后的数与变换前的数只有一个数字不同，并且要保证变换后的四位数也是素数。求最小的变换次数；如果不能完成变换，输出Impossible。
无论怎么变换，个位数字一定是奇数（个位数字为偶数肯定不是素数），这样枚举个位数字时只需枚举奇数就行；而且千位数字不能是0。所以可以用广搜，枚举各个数位上的数字，满足要求的数就加入队列，直到变换成功。因为是广搜，所以一定能保证次数最少。
int n, m;const int N = 1e4 + 100;int vis[N];struct node{ int x, step;};queue<node> Q;bool judge_prime(int x) //判断素数{ if(x == 0 || x == 1) return false; else if(x == 2 || x == 3) return true; else { for(int i = 2; i <= (int)sqrt(x); i++) if(x % i == 0) return false; return true; }}void BFS(){ int X, STEP, i; while(!Q.empty()) { node tmp; tmp = Q.front(); Q.pop(); X = tmp.x; STEP = tmp.step; if(X == m) { printf("%d\n",STEP); return ; } for(i = 1; i <= 9; i += 2) //个位 { int s = X / 10 * 10 + i; if(s != X && !vis[s] && judge_prime(s)) { vis[s] = 1; node temp; temp.x = s; temp.step = STEP + 1; Q.push(temp); } } for(i = 0; i <= 9; i++) //十位 { int s = X / 100 * 100 + i * 10 + X % 10; if(s != X && !vis[s] && judge_prime(s)) { vis[s] = 1; node temp; temp.x = s; temp.step = STEP + 1; Q.push(temp); } } for(i = 0; i <= 9; i++) //百位 { int s = X / 1000 * 1000 + i * 100 + X % 100; if(s != X && !vis[s] && judge_prime(s)) { vis[s] = 1; node temp; temp.x = s; temp.step = STEP + 1; Q.push(temp); } } for(i = 1; i <= 9; i++) //千位 { int s = i * 1000 + X % 1000; if(s != X && !vis[s] && judge_prime(s)) { vis[s] = 1; node temp; temp.x = s; temp.step = STEP + 1; Q.push(temp); } } } printf("Impossible\n"); return ;}int main(){ int t, i; scanf("%d",&t); while(t--) { while(!Q.empty()) Q.pop(); scanf("%d%d",&n,&m); memset(vis,0,sizeof(vis)); vis[n] = 1; node tmp; tmp.x = n; tmp.step = 0; Q.push(tmp); BFS(); } return 0;}http://www.hankcs.com/program/cpp/poj-3126-prime-path.html
#define MAX_N 9999 + 16int prime[MAX_N]; // 第i个素数bool is_prime[MAX_N + 1]; //is_prime[i]为真的时候表示i为素数int sieve(const int& n){ int p = 0; fill(is_prime, is_prime + n + 1, true); is_prime[0] = is_prime[1] = false; for (int i = 2; i <= n; ++i) { if (is_prime[i]) { prime[p++] = i; for (int j = 2 * i; j <= n; j += i) { is_prime[j] = false; } } } return p;}int dp[MAX_N];// 将number的倒数第digit位改成changeint get_next(int number, int digit, int change) { switch (digit) { case 0: return number / 10 * 10 + change; case 1: return number / 100 * 100 + number % 10 + change * 10; case 2: return number / 1000 * 1000 + number % 100 + change * 100; case 3: return number % 1000 + change * 1000; } return 0;}///////////////////////////SubMain//////////////////////////////////int main(int argc, char *argv[]){ // 先做一份素数表 sieve(MAX_N); int N; cin >> N; while (N--) { int from, to; cin >> from >> to; // dfs memset(dp, 0x3f, sizeof(dp)); dp[from] = 0; queue<int> q; q.push(from); while (q.size()) { const int current = q.front(); q.pop(); for (int i = 0; i < 4; ++i) { for (int j = 0; j < 10; ++j) { if (i == 3 && j == 0) { // 将第一位改成0是无意义的 continue; } int next = get_next(current, i, j); if (is_prime[next] == false || dp[next] <= dp[current]) { // 不是素数不行，如果到next已经有更小的那也不用这个变换路径了 continue; } dp[next] = dp[current] + 1; q.push(next); } } } cout << dp[to] << endl; } return 0;}