POJ 3126 / Northwestern Europe 2006 Prime Path (数论&BFS)

Prime Path

http://poj.org/problem?id=3126

Time Limit:  1000MS

Memory Limit: 65536K

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

提示：使用一个结构保存数和路径长度。

处理4位数也可以使用字符串，详见这篇文章。

完整代码：

/*0ms,160KB*/
///还可以设计一个保存搜索结果的算法

#include <cstdio>
#include <cstring>
#include <queue>
using namespace std;

int m, n;
bool pr[10000],vis[10000];
struct st
{
	int x, v;
	st(int xx, int vv): x(xx), v(vv) {}
};

inline void init()
{
	for (int i = 2; i < 100; i++)
		if (!pr[i])
			for (int j = i * i; j < 10000; j += i)
                pr[j] = true;
}

inline int bfs()
{
	memset(vis, 0, sizeof(vis));
	queue<st> q;
	q.push(st(m, 0));///起点
	vis[m] = true;
	while (!q.empty())
	{
		st u = q.front();
		q.pop();
		for (int t = 1; t <= 1000; t *= 10)
		{
			int a = (u.x / t) % 10;
			int b = (u.x - a * t);
			for (int j = 0; j < 10; j++)
			{
				int c = b + j * t;///精妙的设计
				if (vis[c] || pr[c] || j == a || (t == 1000 && j == 0))///千位不能是0
                    continue;
				if (c == n)///到终点了
                    return u.v + 1;
				q.push(st(c, u.v + 1));
				vis[c] = true;
			}
		}
	}
	return -1;
}

int main()
{
	init();
	int t;
	scanf("%d", &t);
	while (t--)
	{
		scanf("%d%d", &m, &n);
		int ans = (m != n ? bfs() : 0);
		if (~ans)
			printf("%d\n", ans);
		else
			puts("Impossible");
	}
	return 0;
}