ACM: 广搜+素数  poj 3126 (素数表…

                                                                 
Prime Path

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

 

题意: 内阁大臣要改办公室门牌. 大臣是素数爱好者.门牌每次更改只可以更改一个数字.

        
现在给定一开始的门牌号,和最后的门牌号. 计算出最少的更改次数就可以成功.

解题思路:

                
1. 还是BFS(广搜) , 这题目做烂了.

                
2. 素数判定, 采用打表解决.

 

代码:

#include <cstdio>

#include <iostream>

#include <cstring>

#include <queue>

using namespace std;

#define MAX 100005

int dir[5] = {1,10,100,1000,10000};

struct node

{

    int
num;

    int
dist;

};

bool prime[MAX*10];

bool visit[MAX*10];

node start , end;

void init()

{

  
 memset(prime,false,sizeof(prime));

    int
i,j;

  
 for(i=2;i<10000;i++)

  
   
 for(j=i;i*j<10000;j++)

  
   
   
 prime[i*j]=true;

}

int bfs(node start)

{

  
 queue<node>
qu;

    node
t;

    int result =
0;

    

  
 qu.push(start);

    

    while(
!qu.empty() )

    {

  
   
 t = qu.front();

  
   
 qu.pop();

  
   
 

  
   
 if(t.num == end.num)

  
   
 {

  
   
   
 result = t.dist;

  
   
   
 break;

  
   
 }

  
   
 

  
   
 for(int i = 0; i < 4; ++i)

  
   
 {

  
   
   
 node tt;

  
   
   
 for(int j = 0; j < 10; ++j)

  
   
   
 {

  
   
   
   
 int num1 = j * dir[i];

  
   
   
   
 int num2 = t.num % dir[i+1];

  
   
   
   
 int num3 = t.num % dir[i];

  
   
   
   
 tt.num = t.num - num2 + num1 + num3;

  
   
   
   
 tt.dist = t.dist + 1;

  
   
   
   
 if(tt.num >= 1000
&& tt.num <= 10000
&& !prime[tt.num]
&& !visit[tt.num])

  
   
   
   
 {

  
   
   
   
   
 visit[tt.num] = true;