POJ 3126 Prime Path(BFS 數字處理)

POJ 3126 Prime Path(BFS 數字處理)

題意 給你兩個4位素數a, b 你每次可以改變a的一位數但要求改變後仍為素數 求a至少改變多少次才能變成b

基礎的bfs 注意數的處理就行了 出隊一個數 然後入隊所有可以由這個素數經過一次改變而來的素數 知道得到b

 

#include 
#include 
using namespace std;
const int N = 10000;
int p[N], v[N], d[N], q[N], a, b;

void initPrime()
{
    memset(v, 0 , sizeof(v));
    for(int i = 2; i * i < N; ++i)
        if(!v[i]) for(int j = i; i * j < N; ++j)  v[i * j] = 1;
    for(int i = 2; i < N ; ++i) p[i] = !v[i];
}

int bfs()
{
    int c, t, le = 0, ri = 0;
    memset(v, 0, sizeof(v));
    q[ri++] = a, v[a] = 1, d[a] = 0;
    while(le < ri)
    {
        c = q[le++];
        if( c == b) return d[c];
        for(int i = 1; i < N; i *= 10)
        {
            for(int j = 0; j < 10; ++j)   //把c第i數量級的數改為j
            {
                if(i == 1000 && j == 0) continue;
                t = c / (i * 10) * i * 10 + i * j + c % i;
                if(p[t] && !v[t])
                    v[t] = 1, d[t] = d[c] + 1, q[ri++] = t;
            }
        }
    }
    return -1;
}

int main()
{
    int cas;
    scanf("%d", &cas);
    initPrime();
    while(cas--)
    {
        scanf("%d%d", &a, &b);
        if((a = bfs()) != -1) printf("%d\n", a);
        else puts("Impossible");
    }
    return 0;
}

 

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