poj3126

Prime Path

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

3
1033 8179
1373 8017
1033 1033

Sample Output

6
7
0

Source

Northwestern Europe 2006

素数筛选+bfs。

代码如下：

#include<stdio.h>
#include<string.h>
int prime[10001];
int number[10001];
int step[10001];
int n, m;

void make_prime(void) {
    memset(prime, 1, sizeof (prime));
    prime[0] = prime[1] = 0;
    for (int i = 2; i <= 100; i++)
        if (prime[i])
            for (int j = i * i; j <= 10000; j += i)
                prime[j] = 0;
}

int bfs(void) {
    int front = 0;
    int rear = 0;
    number[rear++] = n;
    front++;
    step[n] = 1;
    while (!step[m] && rear <= front) {
        int tmp = number[front] / 10 * 10;
        for (int i = 1; i < 10; i++) {
            if (prime[tmp + i] && !step[tmp + i]) {
                step[tmp + i] = step[number[front]] + 1;
                number[rear++] = tmp + i;
            }
        }
        tmp = number[front] / 100 * 100 + number[front] % 10;
        for (int i = 0; i < 10; i++) {
            if (prime[tmp + i * 10] && !step[tmp + i * 10]) {
                step[tmp + i * 10] = step[number[front]] + 1;
                number[rear++] = tmp + i * 10;
            }
        }
        tmp = number[front] / 1000 * 1000 + number[front] % 100;
        for (int i = 0; i < 10; i++) {
            if (prime[tmp + i * 100] && !step[tmp + i * 100]) {
                step[tmp + i * 100] = step[number[front]] + 1;
                number[rear++] = tmp + i * 100;
            }
        }
        tmp = number[front] % 1000;
        for (int i = 1; i < 10; i++) {
            if (prime[tmp + i * 1000] && !step[tmp + i * 1000]) {
                step[tmp + i * 1000] = step[number[front]] + 1;
                number[rear++] + tmp + i * 1000;
            }
        }
    }
}

int main(void) {
    int c;
    make_prime();
    scanf("%d", &c);
    while (c--) {
        scanf("%d%d", &n, &m);
        bfs();
        if (step[m])
            printf("%d\n", step[m] - 1);
        else
            printf("Impossible\n");
    }
}