Prime Path POJ - 3126

Prime Path POJ - 3126

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

bfs求最短路径问题。分别更改四位数的每一位，然后判断是否为素数，若是， 则入队。

#include<iostream>
#include<cmath>
#include<cstring>
#include<queue>
#include<set>
using namespace std;

int n;
int x,y,ans;
bool vis[10010];
set<int>prime;

struct node{
	int num;
	int step;
};

int isprime(int x)//判断是否为素数
{
	if(x==2||x==3)
		return 1;
	if(x%6!=5&&x%6!=1)
		return 0;
	int i,j;
	for(i=5;i<=sqrt(x);i+=6)
	{
		if(x%i==0||x%(i+2)==0)
		{
			return 0;
		}
	}
	return 1;
}

int judge(int s)
{
	if(prime.find(s)==prime.end())
	{
		return false;
	}
	if(vis[s])
	{
		return false;
	}
	return true;
}

void bfs()
{
	queue<node>q;
	node cur,next;
	cur.num=x;
	cur.step=0;
	q.push(cur);
	vis[x]=true;
	while(!q.empty())
	{
		cur=q.front();
		q.pop();
		int j,i,nx;
		if(cur.num==y)
		{
			ans=cur.step;
			return;
		}
		set<int>::iterator it=prime.begin();//迭代器 
		for(i=1000;i<=9000;i+=1000)//千位 
		{
			nx=i+cur.num%1000;
			if(judge(nx))
			{
				vis[nx]=true;
				next.num=nx;
				next.step=cur.step+1;
				q.push(next);
			}
		}
		for(i=0;i<=9;i++)//百位 
		{
			nx=cur.num/1000*1000+cur.num%100+i*100;
			if(judge(nx))
			{
				vis[nx]=true;
				next.num=nx;
				next.step=cur.step+1;
				q.push(next);
			}
		}
		for(i=0;i<10;i++)//十位 
		{
			nx=cur.num/100*100+cur.num%10+i*10;
			if(judge(nx))
			{
				vis[nx]=true;
				next.num=nx;
				next.step=cur.step+1;
				q.push(next);
			}
		}
		for(i=0;i<10;i++)//个位 
		{
			nx=cur.num/10*10+i;
			if(judge(nx))
			{
				vis[nx]=true;
				next.num=nx;
				next.step=cur.step+1;
				q.push(next);
			}
		}
	}
}

int main()
{
	int i,k=0;
	for(i=1000;i<=9999;i++)
	{
		if(isprime(i))
		{
			prime.insert(i);//素数打表 
		}
	}
	int t;
	cin>>t;
	while(t--)
	{
		memset(vis,0,sizeof(vis));
		ans=0;
		cin>>x>>y;
		bfs();
		if(ans)
		{
			cout<<ans<<endl;
		}
		else
		{
			cout<<0<<endl;
		}
	}
	return 0;
}