POJ 3126 Prime Path （BFS）

[align=center]Prime Path[/align]

Time Limit: 1000MS		Memory Limit: 65536K
Total Submissions: 14799		Accepted: 8345

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

题目链接：http://poj.org/problem?id=3126

题目大意：给定两个四位数的质数a,b，由a开始，每次变换一位上的数字得到一个新的四位质数，直到变换得到b,求最少变换的次数。如果没有解，输出“Impossible”。

解题思路：典型BFS暴力，预处理一下素数筛就行，改变位上的数字时用数组存储该四位数。

代码如下：

#include <cstdio>
#include <cstring>
#include <queue>
using namespace std;
int n,m,prm[100000],vis[100000];
struct node
{
int x,cnt;
};
void get_prime()    //得到四位数的素数晒
{
memset(prm,0,sizeof(prm));
for(int i=1000;i<=9999;i++)
{
bool p=true;
for(int j=2;j*j<=i;j++)
{
if(i%j==0)
{
p=false;
break;
}
}
if(p)
prm[i]=1;
}
}
int bfs()
{
queue<node>q;
node sd;
sd.x=n,sd.cnt=0;
q.push(sd);
while(!q.empty())
{
node nn,nd=q.front();
q.pop();
if(nd.x==m)
return nd.cnt;
int a[4];
int tm=nd.x;
for(int i=0;i<4;i++)
{
a[i]=tm%10;
tm/=10;
}
for(int i=0;i<4;i++)         //枚举每一位上的数字
{
for(int j=0;j<=9;j++)        //枚举当前位上要改变的数字
{
if(j==a[i] || i==3&&j==0)continue;
int cur;
if(i==0)
cur=a[3]*1000+a[2]*100+a[1]*10+j;
else if(i==1)
cur=a[3]*1000+a[2]*100+j*10+a[0];
else if(i==2)
cur=a[3]*1000+j*100+a[1]*10+a[0];
else
cur=j*1000+a[2]*100+a[1]*10+a[0];
if(vis[cur] || !prm[cur])continue;   //如果下一个数字被访问过或者不是质数，跳过
vis[cur]=1;
nn.x=cur;
nn.cnt=nd.cnt+1;
q.push(nn);
}
}
}
return -1;
}
int main(void)
{
int t;
get_prime();
scanf("%d",&t);
while(t--)
{
memset(vis,0,sizeof(vis));
scanf("%d%d",&n,&m);
vis
=1;
int ans=bfs();
if(ans==-1)
printf("Impossible\n");
else
printf("%d\n",ans);
}
}