poj3126——Prime Path（BFS）-白红宇

poj3126——Prime Path（BFS）

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发布时间：2019-05-10

本文共 2638 字，大约阅读时间需要 8 分钟。

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

1033 8179

1373 8017

1033 1033

Sample Output

一个四位素数变化成另一个四位素数，每次只能改变一位且改变后的数字也是素数，求最少需要改变多少次。

放在搜索分类下我才知道要用BFS。。。

#include 
   
    #include 
    
     #include 
     
      #include 
      
       #include 
       
        #include 
        #include 
         
          #define INF 0x3f3f3f3f#define MAXN 1010using namespace std;struct Node{ int a[4],step;};int primer[3000],cnt,vis[10000];bool flag;bool judge(int n){ for(int i=2; i*i<=n; ++i) if(n%i==0) return false; return true;}void bfs(Node start,int m){ queue
          
            q; q.push(start); while(!q.empty()) { Node tmp=q.front(),tmp1; q.pop(); int mm=tmp.a[0]*1000+tmp.a[1]*100+tmp.a[2]*10+tmp.a[3]; //cout<<"mm="<
           
            <

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