POJ-3126 Prime Path （BFS数组）

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搜索遍历各个位加入数组即可

#include<iostream>
#include<stdio.h>
#include<string.h>
#include<cmath>
using namespace std;
bool prime[10010];
void is_prime(){
	for(int i=2;i<=9999;i++){
		prime[i]=true;
	}
	for(int i=2;i*i<=10000;i++){
		for(int j=i*i;j<10000;j+=i){
			prime[j]=false;
		}
	}
}
int dp[10010];
int c[5][2]={{1,-1},{10,-10},{100,-100},{1000,-1000},{10000,-10000}};
int n,first,last;
void bfs(int f,int l){
	int q[10010];
	memset(dp,0,sizeof(dp));
	int p=0,r=0;
	q[p]=f;
	dp[f]=1;
	while(p<=r){
		int x=q[p];
		for(int i=0;i<4;i++){
			for(int j=0;j<2;){
				int t=x+c[i][j];
				int y1=(x/c[i+1][0])*c[i+1][0];
				int y2=(x/c[i+1][0])*c[i+1][0]+c[i+1][0];
				while(abs(x-t)<c[i+1][0]&&t>=y1&&t<=y2&&t>=1000&&t<=9999){
					//cout<<"t="<<t<<endl;
					if(!dp[t]&&prime[t]){
						r++;
					    q[r]=t;
					    dp[t]=dp[x]+1;
					    if(t==l){
					    	//flag=1;
					    	printf("%d\n",dp[t]-1);//res=dp[t]-1;
					    	return;
						}
					}
					t+=c[i][j];	
				}
				j++;
			}
		}
		p++;	
		}
		return;
	}
int main(){
	is_prime();
	scanf("%d",&n);
	for(int i=1;i<=n;i++){
		scanf("%d%d",&first,&last);
		if(first==last){
			cout<<"0"<<endl;
		}else{
			  bfs(first,last);	
			}
		}
		return 0;
	}