poj3126_Prime Path

Prime Path

Time Limit: 1000MS Memory Limit: 65536KTotal Submissions: 9833 Accepted: 5648

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

31033 81791373 80171033 1033

Sample Output

670分析：简单广搜+判断素数。
#include <iostream>#include <cstring>#include <cstdio>#include <queue>using namespace std;bool visit[10000];bool prime[10000];int x, y, f;int dd[] = {1000, 100, 10, 1};struct Data {int num;int sum;};queue<Data>q;void fun()//筛法求素数{prime[1] = 1;prime[2] = 0;for (int i = 2; i <= 100; i++){if (!prime[i]){int j = i * 2;while (j < 10000){prime[j] = 1;j = j + i;}}}}int main(){int n;scanf("%d", &n);fun();while (n--){f = 0;scanf("%d%d", &x, &y);memset(visit, 0, sizeof(visit));if (x == y){printf("0\n");continue;}Data data;data.num = x;data.sum = 0;q.push(data);visit[x] = 1;int b[4];while (!q.empty()){data = q.front();q.pop();b[0] = data.num / 1000;   b[1] = data.num / 100 % 10;   b[2] = data.num / 10 % 100 - b[1] * 10;   b[3] = data.num % 1000 - b[1] * 100 - b[2] * 10;//把四位数分离for (int i = 0; i < 4; i++){int l = 1;for (int k = (b[i] + 1) % 10; l <= 9; l++, k = (k + 1) % 10)//变化其中一位{int sum = 0;for (int j = 0; j < 4; j++){if (j == i)sum += k * dd[i];else sum += b[j] * dd[j];}//求和if (sum / 1000 == 0)continue;if (sum == y){printf("%d\n", data.sum + 1);f = 1;break;}if (!visit[sum] && !prime[sum]){Data c;c.num = sum;c.sum = data.sum + 1;visit[sum] = 1;q.push(c);}}if (f == 1)break;}if (f == 1)break;}while (!q.empty())q.pop();}return 0;}