HDU 1973 Prime Path

Prime Path

Time Limit: 5000/1000 MS (Java/Others)    Memory Limit: 65536/32768 K (Java/Others)
Total Submission(s): 763    Accepted Submission(s): 473

Problem 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

题目大意，给你两个素数，要求每次只改变第一个素数的一个数字，问最少需要多少步才能变为第二个素数

很简单，傻瓜广搜呗~

#include
#include
#include
#include
#include
using namespace std;
int a,b;
bool book[10000];
struct Node{
	int num,step;
};
queue que;
bool isprime(int x){
	for(int i=2;i<=sqrt(x);i++){
		if(x%i==0) return false;
	}
	return true;
}
int bfs(){
	Node start;
	start.num=a;
	start.step=0;
	que.push(start);
	while(!que.empty())
	{
		Node now;
		now=que.front();
		que.pop();
		if(now.num==b) return now.step;
		for(int i=1;i<=9;i+=2){
			int x=now.num/10*10+i;
			if(x!=now.num&&!book[x]&&isprime(x)){
				Node next;
				next.num=x;
				next.step=now.step+1;
				book[x]=true;
				que.push(next);
			}
		}
		for(int i=0;i<=9;i++){
			int x=now.num/100*100+i*10+now.num%10;
			if(x!=now.num&&!book[x]&&isprime(x)){
				Node next;
				next.num=x;
				next.step=now.step+1;
				book[x]=true;
				que.push(next);
			}
		}
		for(int i=0;i<=9;i++){
			int x=now.num/1000*1000+i*100+now.num%100;
			if(x!=now.num&&!book[x]&&isprime(x)){
				Node next;
				next.num=x;
				next.step=now.step+1;
				book[x]=true;
				que.push(next);
			}
		}
		for(int i=1;i<=9;i++){
			int x=i*1000+now.num%1000;
			if(x!=now.num&&!book[x]&&isprime(x)){
				Node next;
				next.num=x;
				next.step=now.step+1;
				book[x]=true;
				que.push(next);
			}
		}
	}
}
int main(){
	int t;
	scanf("%d",&t);
	while(t--){
		while(!que.empty()) que.pop();
		memset(book,false,sizeof(book));
		scanf("%d%d",&a,&b);
		int ans=bfs();
		printf("%d\n",ans);
	}
}