poj_3126Prime Path 原

N3verL4nd

Prime Path
 Time Limit: 1000MS Memory Limit: 65536K Total Submissions: 8725 Accepted: 4960

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``````

``````#include <iostream>
#include <cstring>
#include <queue>
using namespace std;
#define MAX 9999

typedef struct Node
{
int x;
int cnt;
}Node;
int primer[MAX];
bool visited[MAX];
queue<Node>Q;
int y;

void pri() //1代表素数
{
for(int i = 1; i < MAX; i++)
{
primer[i] = 1;
}
primer[0] = 0;
primer[1] = 0;
for(int i = 2; i < MAX / 2; i++)
{
for(int j = 2; i * j < MAX; j++)
{
primer[i*j] = 0;
}
}
}

int bfs(int x)
{
int a, b, c;
while (!Q.empty())
{
Q.pop();
}
Node pre, next;
pre.x = x;
pre.cnt = 0;
visited[pre.x] = true;
Q.push(pre);
while (!Q.empty())
{
pre = Q.front();
Q.pop();
if(pre.x == y)
{
//cout << "YES" << endl;
return pre.cnt;
}

c = pre.x / 10; //个位
for(int i = 0; i <= 9; i++)
{
if(primer[c * 10 + i] == 1 && c * 10 + i != pre.x && !visited[c * 10 + i])
{
visited[c * 10 + i] = true;
next.x = c * 10 + i;
next.cnt = pre.cnt + 1;
Q.push(next);
}
}

a = pre.x % 10; //十位
b = pre.x / 100;
for(int i = 0; i <= 9; i++)
{
if(primer[(b * 10 + i) * 10 + a] == 1 && (b * 10 + i) * 10 + a != pre.x && !visited[(b * 10 + i) * 10 + a])
{
visited[(b * 10 + i) * 10 + a] = true;
next.x = (b * 10 + i) * 10 + a;
next.cnt = pre.cnt + 1;
Q.push(next);
}
}

a = pre.x % 100;  //百位
b = pre.x / 1000;
for(int i = 0; i <= 9; i++)
{
if(primer[100*(b * 10 + i) + a] == 1 && 100 * (b * 10 + i) + a != pre.x && !visited[100 * (b * 10 + i) + a])
{
visited[100 * (b * 10 + i) + a] = true;
next.x = 100 * (b * 10 + i) + a;
next.cnt = pre.cnt + 1;
Q.push(next);
}
}

a = pre.x % 1000; //千位
for(int i = 1; i <= 9; i++)
{
if(primer[i * 1000 + a] == 1 && i * 1000 + a != pre.x && !visited[i * 1000 + a])
{
visited[i * 1000 + a] = true;
next.x = i * 1000 + a;
next.cnt = pre.cnt + 1;
Q.push(next);
}
}
}
return 0;
}

int main()
{
pri();
int x, t;
cin >> t;
while (t--)
{
memset(visited, false, sizeof(visited));
cin >> x >> y;
cout << bfs(x) << endl;
}
}``````

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