F - Prime Path POJ - 3126

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

VOJ里C++编译器sqrt里面不可以放int、不可以用to_string、stoi。CE了好几次

#include <iostream>
#include <vector>
#include <algorithm>
#include <string>
#include <set>
#include <queue>
#include <map>
#include <sstream>
#include <cstdio>
#include <cstring>
#include <numeric>
#include <cmath>
#include <iomanip>
//#include <unordered_set>
//#include <unordered_map>
//#include <xfunctional>
#define ll  long long
#define PII  pair<int, int>
using namespace std;
int dir[5][2] = { {0,1} ,{0,-1}, {1,0}, {-1,0} ,{0,0} };
const long long INF = 0x7f7f7f7f7f7f7f7f;
const int inf = 0x3f3f3f3f;
const double pi = 3.14159265358979;
const int mod = 1e9 + 7;
const int maxn = 1e5;
//if(x<0 || x>=r || y<0 || y>=c)
struct node
{
    int cnt, val;
};
int replace(int num,int n,int token)
{
    int arr[4];
    arr[0] = num / 1000;
    arr[1] = (num / 100)%10;
    arr[2] = (num / 10)%10;
    arr[3] = num % 10;
    arr[n] = token;
    int res=arr[0]*1000+arr[1]*100+arr[2]*10+arr[3];
    return res;
}
bool isprime(int num)
{
    if (num == 2 || num == 3)
        return 1;
    if (num % 6 != 1 && num % 6 != 5)
        return 0;
    double nums = num;
    int tmp = sqrt(nums);
    for (int i = 5; i <= tmp; i += 6)
        if (num %i == 0 || num % (i + 2) == 0)
            return 0;
    return 1;
}
int main()
{
    int t;
    cin >> t;
    int visited[maxn];
    while (t--)
    {
        memset(visited, 0, sizeof(visited));
        int a, b;
        cin >> a >> b;
        queue<node> que;
        node a1;
        a1.cnt = 0;
        a1.val = a;
        que.push(a1);
        visited[a] = 1;
        while (!que.empty())
        {
            node front = que.front();
            que.pop();
            if (front.val == b)
            {
                cout << front.cnt<<endl;
                break;
            }
            for (int n = 0; n < 4; n++)
            {
                for (int i = 0; i < 10; i++)
                {
                    if (n == 0 && i == 0)
                        continue;
                    int tmp = replace(front.val, n, i);
                    if (visited[tmp]==0)
                    {
                        visited[tmp] = 1;
                        if (isprime(tmp))
                        {
                            node t;
                            t.cnt = front.cnt + 1;
                            t.val = tmp;
                            que.push(t);
                        }
                    }
                }
            }
        }
    }
    return 0;
}