poj 3126 Prime Path（搜索专题）

Prime Path

Time Limit: 1000MS		Memory Limit: 65536K
Total Submissions: 20237		Accepted: 11282

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

Source

Northwestern Europe 2006

方法很简单，但是我写代码还是写了好久，将思路转化成代码的速度还是太慢。

其中需要素数，又复习了一下素数筛法的方法。

 

 #include <iostream>
 #include <cstdio>
 #include <cstring>
 #include <algorithm>
 #include <cmath>

 using namespace std;

 bool prime[];
 int que[];
 int step[];
 int isprime[];
 bool vis[];
 int cou=;

 void getprime(){
     int num=;
     num=sqrt(num*1.0);
     memset(prime,true,sizeof(prime));
     prime[]=prime[]=false;
     for(int i=;i<;i+=){
         prime[i]=false;
     }
     for(int i=;i<num;i+=){
         if(prime[i]){
             for(int j=i*i;j<;j+=*i){
                 prime[j]=false;
             }
         }
     }
     for(int i=;i<;i++){
         if(prime[i]){
             isprime[cou++]=i;
         }
     }
 }

 bool judge(int a,int b){
     int sum=;
     int t1[],t2[];
     for(int i=;i<;i++){
         t1[i]=a%;
         a/=;
         t2[i]=b%;
         b/=;
     }
     for(int i=;i<;i++){
         if(t1[i]==t2[i]){
             sum++;
         }
     }
     if(sum==){
         return true;
     }else{
         return false;
     }
 }

 int bfs(int a,int b){
     int start=;
     int endd=;
     memset(vis,true,sizeof(vis));
     que[endd]=a;
     step[endd++]=;
     while(start<endd){
         int now=que[start];
         int nowStep=step[start++];
         if(now==b){
             return nowStep;
         }
         for(int j=;j<=cou;j++){
             int i=isprime[j];
             if(!vis[i]){
                 continue;
             }
             if(prime[i]&&judge(i,now)){
                 if(i==b){
                     return nowStep+;
                 }
                 que[endd]=i;
                 step[endd++]=nowStep+;
                 vis[i]=false;
             }
         }
     }
     return ;
 }

 int main()
 {
     int n;
     scanf("%d",&n);
     int a,b;
     getprime();
     for(int i=;i<n;i++){
         scanf("%d %d",&a,&b);
         int ans=bfs(a,b);
         printf("%d\n",ans);
     }
     return ;
 }