poj3126 Prime Path

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.

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).

One line for each case, either with a number stating the minimal cost or containing the word Impossible.

3
1033 8179
1373 8017
1033 1033

6
7
0

题意是给你两个四位数的素数n,m，没有前导0，让你每次变化一位数，使得n变成m，问最小的次数。这题是典型的bfs.因为是4位数，所以判断素数的时候只要判断能不能被2~100整除就行了。

#include<stdio.h>
#include<string.h>
int num,n,m;
int a[10];
int q[1111111][2],vis[50000];    //q[][0]数字,q[][1]次数 
int prime(int n)
{
	int i,flag=1;
	for(i=2;i<=100;i++){
		if(n%i==0){
			flag=0;break;
		}
	}
	if(flag)return 1;
	else return 0;
}


int bfs()
{
	int front=1,rear=1,x,t,i,xx,y;
	memset(vis,0,sizeof(vis));
	vis[n]=1;
	q[front][0]=n;q[front][1]=0;
	while(front<=rear)
	{
		x=q[front][0];
		if(x==m) return q[front][1];
		front++;
		vis[x]=1;
		t=0;
		xx=x;
		while(xx){
			a[++t]=xx%10;
			xx=xx/10;
		}
		
		for(i=1;i<=9;i++){
			y=i*1000+a[3]*100+a[2]*10+a[1];
			if( vis[y]==0 && prime(y)){
				vis[y]=1;
				rear++;q[rear][0]=y;q[rear][1]=q[front-1][1]+1;
			}
		}
		
		for(i=0;i<=9;i++){
			y=i*100+a[4]*1000+a[2]*10+a[1];
			if( vis[y]==0 && prime(y)){
				vis[y]=1;
				rear++;q[rear][0]=y;q[rear][1]=q[front-1][1]+1;
			}
		}
		
		for(i=0;i<=9;i++){
			y=a[4]*1000+a[3]*100+i*10+a[1];
			if( vis[y]==0 && prime(y)){
				vis[y]=1;
				rear++;q[rear][0]=y;q[rear][1]=q[front-1][1]+1;
			}
		}
		
		for(i=0;i<=9;i++){
			y=a[4]*1000+a[3]*100+a[2]*10+i;
			if( vis[y]==0 && prime(y)){
				vis[y]=1;
				rear++;q[rear][0]=y;q[rear][1]=q[front-1][1]+1;
			}
		}
	}
}




int main()
{
	int T,i,j;
	scanf("%d",&T);
	while(T--)
	{
		scanf("%d%d",&n,&m);
		printf("%d\n",bfs());
	}
	return 0;
}