【题解】POJ-3126 Prime Path

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POJ-3126 Prime Path 类型 BFS

Prime Path (POJ - 3126)

题面

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.

输入

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).

输出

One line for each case, either with a number stating the minimal cost or containing the word Impossible.

样例输入

1
2
3
4
3
1033 8179
1373 8017
1033 1033

样例输出

1
2
3
6
7
0

提示

无

思路

欧拉筛打表，枚举每一位的变化，注意大于2的偶数不是质数

代码

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using namespace std;
typedef long long ll;
const int inf = 0x3f3f3f3f;
const int N = 1e5+5;

bool isp[N] = {0};
bool vis[N] = {0};
int n, s, e;

struct P {
    int x;
    int step;
};

void euler() {
    int p[N], m=0;
    for(int i=2; i<=N; i++) {
        if(!isp[i])
            p[m++] = i;
        for(int j=0; j<m; j++) {
            if(p[j]*i>N)
                break;
            isp[p[j]*i] = 1;
            if(i%p[j]==0)
                break;
        }
    }
}

int bfs() {
    memset(vis, 0, sizeof(vis));

    P sp;
    sp.x = s;
    sp.step = 0;

    queue<P> q;
    q.push(sp);
    vis[sp.x] = 1;

    while(!q.empty()) {
        P tp = q.front();
        q.pop();

        if(tp.x==e) {
            return tp.step;
        }
        P np = tp;
        np.step = tp.step+1;

        for(int i=1; i<=9; i+=2) {
            np.x = tp.x/10*10 + i;
            if(np.x!=tp.x && !vis[np.x] && !isp[np.x]) {
                q.push(np);
                vis[np.x] = 1;
            }
        }
        for(int i=0; i<=9; i++) {
            np.x = tp.x/100*100 + i*10 + tp.x%10;
            if(np.x!=tp.x && !vis[np.x] && !isp[np.x]) {
                q.push(np);
                vis[np.x] = 1;
            }
        }
        for(int i=0; i<=9; i++) {
            np.x = tp.x/1000*1000 + i*100 + tp.x%100;
            if(np.x!=tp.x && !vis[np.x] && !isp[np.x]) {
                q.push(np);
                vis[np.x] = 1;
            }
        }
        for(int i=1; i<=9; i++) {
            np.x = i*1000 + tp.x%1000;
            if(np.x!=tp.x && !vis[np.x] && !isp[np.x]) {
                q.push(np);
                vis[np.x] = 1;
            }
        }
    }
    return 0;
}

int main(void) {
    euler();
    scanf("%d", &n);
    for(int i=0; i<n; i++) {
        scanf("%d%d", &s, &e);
        printf("%d\n", bfs());
    }
    return 0;
}