poj3126--Prime Path(广搜)

Prime Path

Time Limit: 1000MS		Memory Limit: 65536K
Total Submissions: 11751		Accepted: 6673

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

给出两个数s，e，都是素数。经过转化，将s转化为e的最小步数

规则，每一次仅仅能改动一位，每次得到的数都是素数。

素数筛跑出1000到10000内的全部素数。假设当中两个素数仅仅有一位不同。那么连接一条边。得到全部素数组合的图后用bfs直接搜索就能够

#include <cstdio>
#include <cstring>
#include <queue>
#include <algorithm>
using namespace std;
int a[11000] , check[11000] , tot ;
struct node{
    int v , next ;
}p[2000000];
struct node1{
    int u , t ;
};
queue <node1> que ;
int head[10000] , cnt , flag[10000] ;
void add(int u,int v)
{
    p[cnt].v = v ;
    p[cnt].next = head[u] ;
    head[u] = cnt++ ;
}
void init()
{
    memset(check,0,sizeof(check));
    memset(head,-1,sizeof(head));
    tot = cnt = 0 ;
    int i , j , k , num ;
    for(i = 2 ; i <= 10000 ; i++)
    {
        if( !check[i] )
            a[tot++] = i ;
        for(j = 0 ; j < tot ; j++)
        {
            if(i*a[j] >= 10000)
                break;
            check[i*a[j]] = 1 ;
            if( i%a[j] == 0 )
                break;
        }
    }
    for(i = 0 ; i < tot ; i++)
        if( (a[i]/1000) ) break;
    k = i ;
    for(i = k ; i < tot ; i++)
    {
        for(j = k ; j < i ; j++)
        {
            num = 0 ;
            if( a[i]%10 != a[j]%10 )
                num++ ;
            if( a[i]/10%10 != a[j]/10%10 )
                num++ ;
            if( a[i]/100%10 != a[j]/100%10 )
                num++ ;
            if( a[i]/1000%10 != a[j]/1000%10 )
                num++ ;
            if(num == 1)
            {
                add(i,j);
                add(j,i);
            }
        }
    }
}
int find1(int x)
{
    int low = 0 , mid , high = tot-1 ;
    while(low <= high)
    {
        mid = (low+high)/2 ;
        if(a[mid] == x)
            return mid ;
        else if(a[mid] < x)
            low = mid + 1 ;
        else
            high = mid -1 ;
    }
}
int bfs(int s,int e)
{
    memset(flag,0,sizeof(flag));
    while( !que.empty() )
        que.pop();
    int i , j , v ;
    node1 low , high ;
    low.u = s ;
    low.t = 0 ;
    flag[s] = 1 ;
    que.push(low);
    while( !que.empty() )
    {
        low = que.front();
        que.pop();
        if( low.u == e )
            return low.t ;
        for(i = head[low.u] ; i != -1 ; i = p[i].next)
        {
            v = p[i].v ;
            if( !flag[v] )
            {
                flag[v] = 1;
                high.u = v ;
                high.t = low.t + 1 ;
                que.push(high);
            }
        }
    }
    return 0;
}
int main()
{
    int t , s , e ;
    init();
    scanf("%d", &t);
    while(t--)
    {
        scanf("%d %d", &s, &e);
        s = find1(s);
        e = find1(e);
        printf("%d\n", bfs(s,e) );
    }
    return 0;
}