POJ 3126 Prime Path
Description — 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.
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 Sample Output 6 大致题意： 给定两个四位素数a b，要求把a变换到b 变换的过程要保证 每次变换出来的数都是一个 四位素数，而且当前这步的变换所得的素数 与 前一步得到的素数 只能有一个位不同，而且每步得到的素数都不能重复。 求从a到b最少需要的变换次数。无法变换则输出Impossible 注意：双向广搜是在一个队列中实现的，只不过是交替进行罢了！ |
#include<iostream>
using namespace std;
#include<cstdio>
#include<cstring>
#include<cmath>
#include<cstdlib>
#include<queue>
#define N 10000
struct prime{
int c[];
int flag;
};
int dis[N];
int visit[N];
queue<prime>que;
int js(int *m)
{
return (m[]*+m[]*+m[]*+m[]*);
}
bool is_prime(int l)
{
bool flag=true;
for(int i=;i<=sqrt(l);++i)
{
if(l%i==)
{
flag=false;
break;
}
}
return flag;
}
int bfs()
{
while(!que.empty())
{
prime x=que.front();
que.pop();
int now=js(x.c);
for(int i=;i<=;++i)
{
prime nx=x;
nx.c[]=i;
int shu=js(nx.c);
if(!visit[shu]&&is_prime(shu))
{
visit[shu]=x.flag;
nx.flag=x.flag;
que.push(nx);
dis[shu]=dis[now]+;
}
else if(visit[shu]&&visit[shu]!=x.flag)
{
return dis[now]+dis[shu]+;
}
}
for(int j=;j<=;++j)
{
for(int i=;i<=;++i)
{
prime nx=x;
nx.c[j]=i;
int shu=js(nx.c);
if(!visit[shu]&&is_prime(shu))
{
visit[shu]=x.flag;
nx.flag=x.flag;
que.push(nx);
dis[shu]=dis[now]+;
}
else if(visit[shu]&&visit[shu]!=x.flag)
{
return dis[now]+dis[shu]+;
}
} }
}
return -;
}
int main()
{
int tex;
scanf("%d",&tex);
char a[];
while(tex--)
{
while(!que.empty()) que.pop();
memset(dis,,sizeof(dis));
memset(visit,,sizeof(visit));
scanf("%s",a);
que.push(prime{a[]-'',a[]-'',a[]-'',a[]-'',});
int p=(a[]-'')*+(a[]-'')*+(a[]-'')*+(a[]-'');
visit[p]=;dis[p]=;
int q=p;
scanf("%s",a);
que.push(prime{a[]-'',a[]-'',a[]-'',a[]-'',});
p=(a[]-'')*+(a[]-'')*+(a[]-'')*+(a[]-'');
visit[p]=;dis[p]=;
if(q==p)
{
printf("0\n");continue;
}
int temp=bfs();
if(temp==-) printf("Impossible\n");
else printf("%d\n",temp);
}
return ;
}