POJ3126 Prime Path【BFS、埃氏筛】

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

解题思路

题目大意：给出两个四位无前导0素数，定义一次变化为改变一个数位上的数字，且变化后的数仍为四位无前导0素数。问这两个素数间需要经过几次变化。

如：1033-1733-3733-3739-3779-8779-8179，计6步。

解题思路：首先，我们先使用埃氏筛将所有四位数中的素数打表记录。然后，从起始数开始，使用BFS对每个数改变各数位上的值，然后直到我们找到了终点数为止。

#include<iostream>
#include<cstdio>
#include<algorithm>
#include<queue>
#include<cstring>
using namespace std;
const int MAXN = 10010;
int start, en;                                                //起点和终点
int prime[MAXN], p[MAXN] = { 0 }, pNum = 0;      //如果i为素数则p[i] = 0;  prime[]中保存了所有四位数素数
int vis[MAXN] = { 0 };

struct Point
{
	int num;
	int step;
};

void find_Prime()              //埃氏筛，寻找四位数的素数
{
	for (int i = 2; i < 10000; i++)       
	{
		if (p[i] == 0)
		{
			if (i > 999) prime[pNum++] = i;
			for (int j = i + i; j < 10000; j += i)
			{
				p[j] = 1;
			}
		}
	}
}

int BFS(int x)
{
	queue<Point> point;
	Point s = { x,0 };
	point.push(s);
	vis[x] = 1;
	while (!point.empty())      //非空
	{
		Point tmp = point.front();
		point.pop();
		int num[4];
		int t = 0;
		while (tmp.num)
		{
			num[t++] = tmp.num % 10;
			tmp.num /= 10;
		}
		int nextNum;
		for (int i = 1; i <= 9; i++)                                    //开始遍历每一位数
		{
			nextNum = i * 1000 + num[2] * 100 + num[1] * 10 + num[0];
			if (vis[nextNum] == 0 && p[nextNum]==0)                                 //如果没有访问过，且该数为素数
			{
				Point next;
				next.num = nextNum;
				next.step = tmp.step + 1;
				vis[nextNum] = 1;
				point.push(next);
				if (nextNum == en)   //如果下一个数是终点
					return next.step;
			}
		}
		for (int i = 0; i <= 9; i++)                                    //开始遍历每一位数
		{
			nextNum = num[3] * 1000 + i * 100 + num[1] * 10 + num[0];
			if (vis[nextNum] == 0 && p[nextNum]==0)                                 //如果没有访问过，且该数为素数
			{
				Point next;
				next.num = nextNum;
				next.step = tmp.step + 1;
				vis[nextNum] = 1;
				point.push(next);
				if (nextNum == en)   //如果下一个数是终点
					return next.step;
			}
		}
		for (int i = 0; i <= 9; i++)                                    //开始遍历每一位数
		{
			nextNum = num[3] * 1000 + num[2] * 100 + i * 10 + num[0];
			if (vis[nextNum] == 0 && p[nextNum]==0)                                 //如果没有访问过，且该数为素数
			{
				Point next;
				next.num = nextNum;
				next.step = tmp.step + 1;
				vis[nextNum] = 1;
				point.push(next);
				if (nextNum == en)   //如果下一个数是终点
					return next.step;
			}
		}
		for (int i = 0; i <= 9; i++)                                    //开始遍历每一位数
		{
			nextNum = num[3] * 1000 + num[2] * 100 + num[1] * 10 + i;
			if (vis[nextNum] == 0 && p[nextNum]==0)                                 //如果没有访问过，且该数为素数
			{
				Point next;
				next.num = nextNum;
				next.step = tmp.step + 1;
				vis[nextNum] = 1;
				point.push(next);
				if (nextNum == en)   //如果下一个数是终点
					return next.step;
			}
		}
	}
	return 0;
}


int main()
{
	int t;
	scanf("%d", &t);
	find_Prime();
	while (t--)
	{
		scanf("%d%d", &start, &en);
		memset(vis, 0, sizeof(vis));
		if (start == en)
			printf("0\n");
		else
		{
			int ans = BFS(start);
			if (ans == 0)
				printf("Impossible\n");
			else
				printf("%d\n", ans);
		}
	}
	return 0;
}