题目链接：http://acm.hdu.edu.cn/showproblem.php?pid=1973
Prime Path
Time Limit: 5000/1000 MS (Java/Others) Memory Limit: 65536/32768 K (Java/Others)
Problem 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.
— 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
Source
题目大意：给定两个四位素数a b，要求把a变换到b变换的过程要保证 每次变换出来的数都是一个 四位素数，而且当前这步的变换所得的素数
与前一步得到的素数 只能有一个位不同，而且每步得到的素数都不能重复。求从a到b最少需要的变换次数。无法变换则输出Impossible
解题思路：打一个素数表，然后基于每个数的每一位bfs搜索即可，具体的可见代码~~~
代码如下：
1 #include <iostream> 2 #include <queue> 3 #include <cstdio> 4 #include <cstring> 5 using namespace std; 6 struct node{ 7 int cur, step; 8 }now, Next; 9 int vis[10001], star, finish, prime[10001] = { 1, 1, 0 }; 10 void init(){ 11 for (int i = 2; i < 10001; i++){ 12 if (!prime[i]){ 13 for (int j = 2; i*j < 10001; j++) 14 prime[i*j] = 1; 15 } 16 } 17 } 18 int bfs(){ 19 queue<node> Q; 20 vis[star] = 1; 21 now.cur = star, now.step = 0; 22 Q.push(now); 23 while (!Q.empty()){ 24 int i, j; 25 char num[5]; 26 now = Q.front(); 27 Q.pop(); 28 if (now.cur == finish) return now.step; 29 for (i = 0; i < 4; i++){ 30 sprintf(num, "%d", now.cur); 31 for (j = 0; j < 10; j++){ 32 if (j == 0 && i == 0) 33 continue; 34 if (i == 0) 35 Next.cur = j * 1000 + (num[1] - '0') * 100 + (num[2] - '0') * 10 + (num[3] - '0'); 36 else if (i == 1) 37 Next.cur = j * 100 + (num[0] - '0') * 1000 + (num[2] - '0') * 10 + (num[3] - '0'); 38 else if (i == 2) 39 Next.cur = j * 10 + (num[0] - '0') * 1000 + (num[1] - '0') * 100 + (num[3] - '0'); 40 else if (i == 3) 41 Next.cur = j + (num[0] - '0') * 1000 + (num[1] - '0') * 100 + (num[2] - '0') * 10; 42 if (!prime[Next.cur] && !vis[Next.cur]) 43 { 44 Next.step = now.step + 1; 45 vis[Next.cur] = 1; 46 Q.push(Next); 47 } 48 } 49 } 50 } 51 return -1; 52 } 53 int main(){ 54 int t, ans; 55 cin >> t; 56 init(); 57 while (t--){ 58 cin >> star >> finish; 59 memset(vis, 0, sizeof(vis)); 60 ans = bfs(); 61 if (ans == -1) cout << "Impossible "; 62 else cout << ans << endl; 63 } 64 return 0; 65 }
其实这道题学校OJ(Swust OJ)也有但是坑爹的后台数据变成a+b的后台数据了，Orz~~~(无爱了)