poj_3126 Prime Path

先建素數表。一個數字要改變某位上的數時就轉成字符數組，傳遞參數時再轉成整型。用bfs搜。

 

3126Accepted712K32MSG++1689B2014-07-16 00:30:11

Prime Path

Time Limit: 1000MS Memory Limit: 65536KTotal Submissions: 10968 Accepted: 6236

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

Northwestern Europe 2006

 

代碼：

 

#include<cstdio>#include<cstdlib>#include<cstring>#include<queue>using namespace std;int p[10000],visit[10000],deep[10000];void gen_prime(){   int i,j;   p[1]=1;   for(i=2;i<=5000;i++)   {      j=2;      while(1)      {         int k=i*j;         if(k>10000) break;         p[k]=1;         j++;      }   }}int bfs(int x,int y){   queue<int>q;   char c[5];   int i,j;//   memset(c,'\0',sizeof(c));//   sprintf(c,"%d",x);   if(x==y) return 0;   q.push(x);   visit[x]=1;   while(!q.empty())   {//      printf("en\n");    int h=q.front();    memset(c,'\0',sizeof(c));    sprintf(c,"%d",h);//    printf("d=%s h=%d\n",c,h);    q.pop();    for(i=0;i<10;i++)      {         for(j=0;j<4;j++)         {         if(!(i==0 && j==0))            {             char tmp[5];             memset(tmp,'\0',sizeof(tmp));             strcpy(tmp,c);             tmp[j]=i+'0';             int k=atoi(tmp);//             printf("str=%s k=%d\n",tmp,k);             if(!p[k] && !visit[k])             {               visit[k]=1;               q.push(k);               deep[k]=deep[h]+1;               if(k==y) return deep[k];             }            }         }      }   }   return -1;}int main(){   freopen("in.txt","r",stdin);   int n,m,i,j;   memset(p,0,sizeof(p));   gen_prime();   scanf("%d",&n);   while(n--)   {      int x,y;      scanf("%d%d",&x,&y);      char c[5];      memset(c,'\0',sizeof(c));      memset(visit,0,sizeof(visit));      memset(deep,0,sizeof(deep));      int ans=bfs(x,y);      printf("%d\n",ans);   }   return 0;}