POJ 3126 math（BFS）

Prime Path

Time Limit: 1000MS		Memory Limit: 65536K
Total Submissions: 21581		Accepted: 11986

Descripti

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

思路：对每位数字替换，直到成为目标数字。

代码：

 #include "cstdio"
 #include "iostream"
 #include "algorithm"
 #include "string"
 #include "cstring"
 #include "queue"
 #include "cmath"
 #include "vector"
 #include "map"
 #include "stdlib.h"
 #include "set"
 #define mj
 #define db double
 #define ll long long
 using namespace std;
 const int N=1e4+;
 const int mod=1e9+;
 const ll inf=1e16+;
 bool pri[N];
 bool u[N],v[N];
 int c[N];//统计步数
 void init(){
     int i,j;
     for(i=;i<=N;i++){
         for(j=;j<i;j++)
             if(i%j==){
                 pri[i]=;
                 break;
             }
         if(j==i) pri[i]=;
     }
 }
 int bfs(int s,int e){
     queue<int >q;
     memset(v,, sizeof(v));
     memset(c,, sizeof(c));
     int tmp,a[],ans=s;
     q.push(s);
     v[s]=;
     while(q.size()){
         int k=q.front();
         q.pop();
         a[]=k/,a[]=k/%,a[]=k/%,a[]=k%;
         for(int i=;i<;i++){
             tmp=a[i];
             for(int j=;j<;j++){
                 if(tmp!=j){
                     a[i]=j;
                     ans=a[]*+a[]*+a[]*+a[];
                     if(pri[ans]&&!v[ans]){//为素数且未使用过
                         c[ans]=c[k]+;
                         v[ans]=;
                         q.push(ans);
                     }
                     if(ans==e) {
                         printf("%d\n",c[ans]);
                         return ;
                     }
                 }
                 a[i]=tmp;
             }
         }
         if(k==e) {
             printf("%d\n",c[k]);
             return ;
         }
     }
     printf("Impossible\n");
     return ;
 }
 int main()
 {
     int n;
     int x,y;
     scanf("%d",&n);
     init();
     for(int i=;i<n;i++){
         scanf("%d%d",&x,&y);
         bfs(x,y);
     }
     return ;
 }