poj_3126Prime Path

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

筛选法求出素数表，然后BFS，不加visited数组，poj超时，hdoj溢出，毕竟重复的节点还是蛮多的。

#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;
	}
}