POJ 3126 Prime Path

POJ 3126 Prime Path

题目链接http://poj.org/problem?id=3126

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

题意：

给你两个四位数。从第一个四位数变换到第二个四位数，中间每次只能够变换一位并且中间的每一次变换的结果都是质数，求最短步骤。

题解：

先打质数表，然后bfs求结果。

代码：

#include <iostream>
#include <cstdio>
#include <cstring>
#include <string>
#include <algorithm>
#include <queue>
#include <stack>
#include <map>
using namespace std;
bool prime[11000];
void init()
{
	int key;
	for (int i = 1000;i <= 9999;i++){
		key = 1;
		for (int j = 2;j < i;j++){
			if (i%j == 0){
				key = 0;
				break;
			}
		}
		if (key)
			prime[i] = true;
	}
}
bool vis[10000];
int cot[10000];
int bfs(int a,int b)
{
	memset(vis,0,sizeof(vis));
	memset(cot,0,sizeof(cot));
	queue <int>qu;
	vis[a] = 1;
	qu.push(a);
	int bit[10];
	while (!qu.empty()){
		int temp = qu.front();
		qu.pop();

		if (temp == b)
			return cot[b];
		bit[1] = temp/1000;bit[2] = (temp/100)%10;bit[3] = (temp/10)%10;bit[4] = temp%10;
		for (int bi = 1;bi <= 4;bi ++){
			int temp1 = bit[bi];
			for (int i = 0;i < 10;i++){
				if (bi == 1 && i == 0)
					continue;
				bit[bi] = i;
				int sum = bit[1]*1000+bit[2]*100+bit[3]*10+bit[4];
				if (prime[sum] && (!vis[sum])){
					vis[sum] = 1;
					cot[sum] = cot[temp]+1;
					qu.push(sum);
				}
			}
			bit[bi] = temp1;
		}
	}
	return -1;
}
int main()
{
	init();
	int t,a,b;
	cin>>t;
	while (t--){
		cin>>a>>b;
		int ans = bfs(a,b);
		if (ans == -1)
			cout<<"Impossible"<<endl;
		else cout<<ans<<endl;
	}
	return 0;
}