POJ 3126 Prime Path

原题链接：

http://poj.org/problem?id=3126

题目大意：

先输入一个n（n不超过100）。表示有n组的测试样例。然后输入两个素数。第一个表示起始数。第二个表示目标数。问要多少是操作才可以把第一个数变成第二个数。
操作要求：
1.每次操作后的数必须是素数
2.每次操作只能改动一个数字
如果可以输出需要多少次。不行输出“Impossible”

思路：

采用一种笨办法：
从-9~9每个计算过去小于0或者大于9都舍去。然后判断是不是素数
其他的和一般的BFS类似

代码如下：

#include <iostream>#include <cstdio>#include <cstring>#include <queue>#include <cmath>using namespace std;int T;int Start,End;int Check[10005];int Pre[10005];int number;int Change[18]={-9,-8,-7,-6,-5,-4,-3,-2,-1,1,2,3,4,5,6,7,8,9};bool Find;bool is_prime(int n){    int result = 0;    int i;    if (n <= 1)    {        return result;    }    if(2 == n)    {        result = 1;        return result;    }    for (i = 2; i < n ;i++)    {        if (0 == n % i)        {            result = 0;            break;        }        else if (i + 1 == n)        {            result = 1;        }        else        {            continue;        }    }    return result;}void GetCount( int Index ){    while( Pre[Index] != -1 )    {        Index = Pre[Index];        number++;    }}bool IsNumberLeagal( int x ){    if( x < 0 || x > 9 )        return false;    return true;}bool IsLeagal( int Index ){    if( Index < 1000 || Index > 9999 )        return false;    return true;}void bfs(){    queue<int> Path;    memset( Check,0,sizeof( Check ) );    memset( Pre,0,sizeof( Pre ) );    Check[Start] = 1;    Pre[Start] = -1;    Path.push ( Start );    int Now,Next,fir,sec,thr,fou;    int i,nfir,nsec,nthr,nfou;    while( !Path.empty () )    {        Now = Path.front ();        Path.pop ();        fir = ( Now % 10000 ) /1000;        sec = ( Now % 1000 ) /100;        thr = ( Now % 100 ) /10;        fou = ( Now % 10 ) /1;        for( i = 0; i < 18; i++ )        {            nfir = fir + Change[i];            if( !IsNumberLeagal( nfir ) )                continue;            Next = nfir *1000 + sec *100 + thr *10 + fou *1;            if( IsLeagal( Next ) && !Check[Next] && is_prime( Next ) )            {                Check[Next] = 1;                Pre[Next] = Now;                if( Next == End )                {                    Find = true;                    GetCount( Next );                    return ;                }                Path.push ( Next );            }        }        for( i = 0; i < 18; i++ )        {            nsec = sec + Change[i];            if( !IsNumberLeagal( nsec ) )                continue;            Next = fir *1000 + nsec *100 + thr *10 + fou *1;            if( IsLeagal( Next ) && !Check[Next] && is_prime( Next ) )            {                Check[Next] = 1;                Pre[Next] = Now;                if( Next == End )                {                    Find = true;                    GetCount( Next );                    return ;                }                Path.push ( Next );            }        }        for( i = 0; i < 18; i++ )        {            nthr = thr + Change[i];            if( !IsNumberLeagal( nthr ) )                continue;            Next = fir *1000 + sec *100 + nthr *10 + fou *1;            if( IsLeagal( Next ) && !Check[Next] && is_prime( Next ) )            {                Check[Next] = 1;                Pre[Next] = Now;                if( Next == End )                {                    Find = true;                    GetCount( Next );                    return ;                }                Path.push ( Next );            }        }        for( i = 0; i < 18; i++ )        {            nfou = fou + Change[i];            if( !IsNumberLeagal( nfou ) )                continue;            Next = fir *1000 + sec *100 + thr *10 + nfou *1;            if( IsLeagal( Next ) && !Check[Next] && is_prime( Next ) )            {                Check[Next] = 1;                Pre[Next] = Now;                if( Next == End )                {                    Find = true;                    GetCount( Next );                    return ;                }                Path.push ( Next );            }        }    }}int main(){    cin >> T;    while( T-- )    {        cin >> Start >> End;        if( Start == End )        {            cout<<0<<endl;            continue;        }        number = 0;        Find = false;        bfs();        if( Find )            cout<< number <<endl;        else            cout<< "Impossible" << endl;    }    return 0;}