寒假前刷题（6）广度优先搜索 poj 3126

Prime Path

Time Limit: 1000MS		Memory Limit: 65536K
Total Submissions: 8097		Accepted: 4610

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

//这是一道典型的搜索题，题目的大意是通过每次变换一个数，使得最终能够达到目标的数，每次变换的数都只能是素数，而且变换的过程中数与数之间之能有一个数字不同，这题首先得素数的判定法，比如数m要为素数，则从2到√m，都不能被m整除。其次就是依次把m的每个位的数分离出来，然后进行变换。借鉴了别人的解题报告，所以大神请绕道。

#include<math.h>
#include<queue>
#include<stdio.h>
#include<string.h>
using namespace std;
int a,b;
int p[999999]={0};//存放答案的数组
int visit[999999]={0};//标记数组，标记是否被访问
bool isprime(int x)
{
     int i;
     for(i=2;i<=sqrt((double)x);i++)
     {
          if(x%i==0)
            return false;
     }
     return true;
}
int bfs(int s,int r)
{
     queue<int> q;
     q.push(s);
     p[s]=0;
     visit[s]=1;
     while(!q.empty())
     {
          int temp=q.front();
          q.pop();
          for(int i=0;i<=9;i++)
          {
               int y1=(temp/10)*10+i; //得到个位数
               if(isprime(y1)&&!visit[y1])
               {
                    q.push(y1);
                    visit[y1]=1;
                    p[y1]=p[temp]+1;
               }
               int y2=temp%10+(temp/100)*100+i*10;//得到十位数
               if(isprime(y2)&&!visit[y2])
               {
                    q.push(y2);
                    visit[y2]=1;
                    p[y2]=p[temp]+1;
               }
               int y3=temp%100+(temp/1000)*1000+i*100;//得到百位数
               if(isprime(y3)&&!visit[y3])
               {
                    q.push(y3);
                    visit[y3]=1;
                    p[y3]=p[temp]+1;
               }
               if(i!=0)//这一点一开始我也没注意到，记住如果i==0的话，那么无法分离出最高位。
               {
                    int y4=temp%1000+i*1000;
                    if(isprime(y4)&&!visit[y4])
                    {
                         q.push(y4);
                         visit[y4]=1;
                         p[y4]=p[temp]+1;
                    }
               }
          }
         if(visit[r])
           return p[r];
     }
     return 0;
}
int main()
{
     int T;
     scanf("%d",&T);
     while(T--)
     {
          memset(visit,0,sizeof(visit));
          memset(p,0,sizeof(p));
          scanf("%d %d",&a,&b);
          printf("%d\n",bfs(a,b));
     }
     return 0;
}