poj 3126 Prime Path

Prime Path

Time Limit: 1000MS		Memory Limit: 65536K
Total Submissions: 17651		Accepted: 9941

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

Source
这一个题的题意是
给定两个素数a b，求a变幻到b最少需要几步
并且变幻时只有一个数字不同，并且是素数

比方说上面讲的1033不能变化到8033因为8033不是素数，

解题思想就是
首先要对素数进行打表
然后用bfs一个一个的改变位数的值进行判断就行了，
还要知道的是一个素数到另一个素数是一定能实现的，所以就不用考虑不能实现怎么办的问题了。

代码如下

#include <cstdio>
#include <cstring>
#include <cmath>
#include <iostream>
#include <queue>
#include <algorithm>
using namespace std;

struct node
{
int step;
int x;
};
int Prime[10002];
int repeat[10002];
int t[4];
queue <node> q;

void change(int x)
{
t[0] = x%10;
t[1] = x/10%10;
t[2] = x/100%10;
t[3] = x/1000;
}

void init()
{
int i, j;
int num;
for ( i = 1001; i < 10000;i = i+2 )
{
num = sqrt(i+0.5);
for ( j = 2;j <= num;  j++ )
{
if( i%j == 0 )
{
Prime[i] = 0;
break;
}
}
if ( j > num )
{
Prime[i] = 1;
}
}
}

int bfs( int n, int m )
{
int i;
while( !q.empty() )
{
q.pop();
}
node temp;
temp.x = n;
temp.step = 0;
q.push(temp);
while ( !q.empty() )
{
node tail;
temp = q.front();
q.pop();
change(temp.x);
for ( i = 1;i < 10; i = i+2)
{
tail.x = i+t[1]*10+t[2]*100+t[3]*1000;
if (Prime[tail.x] == 1&&repeat[tail.x] == 0)
{
repeat[tail.x] = 1;
tail.step = temp.step+1;
if ( tail.x == m )
return tail.step;
q.push(tail);
}
}
for ( i = 0;i < 10; i++)
{
tail.x = t[0]+i*10+t[2]*100+t[3]*1000;
if (Prime[tail.x] == 1&&repeat[tail.x] == 0)
{
repeat[tail.x] = 1;
tail.step = temp.step+1;
if ( tail.x == m )
return tail.step;
q.push(tail);
}
}
for ( i = 0;i < 10; i++)
{
tail.x = t[0]+t[1]*10+i*100+t[3]*1000;
if (Prime[tail.x] == 1&&repeat[tail.x] == 0)
{
repeat[tail.x] = 1;
tail.step = temp.step+1;
if ( tail.x == m )
return tail.step;
q.push(tail);
}
}
for ( i = 1;i < 10; i++)
{
tail.x = t[0]+t[1]*10+t[2]*100+i*1000;
if (Prime[tail.x] == 1&&repeat[tail.x] == 0)
{
repeat[tail.x] = 1;
tail.step = temp.step+1;
if ( tail.x == m )
return tail.step;
q.push(tail);
}
}
}
return -1;
}

int main()
{
int n, m, T;
init();
scanf ( "%d", &T );
while ( T-- )
{
memset(repeat, 0, sizeof(repeat));
scanf ( "%d %d", &n, &m );
if ( n == m )
{
printf ("0\n");
continue;
}
int sum = bfs( n, m );
printf ( "%d\n", sum );
}
}

代码菜鸟，如有错误，请多包涵！！！
如有帮助记得支持我一下，谢谢！！！