POJ3126——Prime Path

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

31033 81791373 80171033 1033

Sample Output

670

Source

简单的BFS题，先把10000以内的素数打出来，然后就是普通的BFS做法，枚举修改每一位，反正就4位数

#include<queue>#include<cstdio>#include<cstring>#include<iostream>#include<algorithm>using namespace std;bool is_prime[10010];bool vis[10010];int num[4];struct node{int x;int step;};void get_prime(){for(int i=2;i<=10000;i++)is_prime[i]=1;for(int i=2;i<=10000;i++){if(is_prime[i]){if(10000/i < i)break;for(int j=i*i;j<=10000;j+=i)is_prime[j]=false;}}}void bfs(int s,int e){queue<node>qu;memset(vis,0,sizeof(vis));while(!qu.empty())qu.pop();node tmp1,tmp2;tmp1.x=s;tmp1.step=0;qu.push(tmp1);vis[s]=1;bool flag=false;while(!qu.empty()){tmp1=qu.front();if(tmp1.x==e){printf("%d\n",tmp1.step);flag=true;break;}int y=tmp1.x;qu.pop();num[0]=y/1000;num[1]=y/100%10;num[2]=y/10%10;num[3]=y%10;for(int i=0;i<4;i++)//修改第几位 {for(int j=0;j<=9;j++){if(j==0 && i==0)//修改最高位只能是1-9 continue;if(num[i]!=j){int cnt=1000,newx=0;for(int k=0;k<4;k++){if(k!=i)newx+=num[k]*cnt;elsenewx+=j*cnt;cnt/=10;}//printf("%d\n",newx);if(!vis[newx] && is_prime[newx]){vis[newx]=1;tmp2.x=newx;tmp2.step=tmp1.step+1;qu.push(tmp2);}}   }}}if(!flag)printf("Impossible\n");}int main(){int t;scanf("%d",&t);get_prime();while(t--){int s,e;scanf("%d%d",&s,&e);bfs(s,e);}return 0;}