Sicily 2011. Nine Digits

解题思路是宽搜+康托压缩。具体就是从123456789开始向外宽搜，找到所有情况并记录，然后对输入进行常数时间的查找，最后输出结果。另外，由于1到9的全排列有9!个，不加处理直接开数组的话，可以开10^9（10表示1-10这10个数，9表示有9个位）个，但是这样空间就很大。而利用康托压缩，根据全排列中每个数只会出现一次的实际情况，可以将数组大小从10^9压缩为9!个，也即是去除了重复出现某些数字（缺失某些数字）的排列情况，具体实现请参考代码。
另：简言之，康托压缩是通过判断当前排列在全排列中属于第几大排列而实现的。

Run Time: 0.67sec

Run Memory: 2120KB

Code length: 3177Bytes

SubmitTime: 2012-01-12 01:13:13

// Problem#: 2011
// Submission#: 1188530
// The source code is licensed under Creative Commons Attribution-NonCommercial-ShareAlike 3.0 Unported License
// URI: http://creativecommons.org/licenses/by-nc-sa/3.0/
// All Copyright reserved by Informatic Lab of Sun Yat-sen University
#include <cstdio>
#include <queue>
using namespace std;

int factorial[ 9 ] = { 40320, 5040, 720, 120, 24, 6, 2, 1, 1 };
int tile[ 362880 ] = { };

void rotate1( int from[], int to[] );
void rotate2( int from[], int to[] );
void rotate3( int from[], int to[] );
void rotate4( int from[], int to[] );

void itoa( int n, int a[] ) {
    for ( int i = 8; i >= 0; i-- ) {
        a[ i ] = n % 10;
        n /= 10;
    }
}

int atoi( int a[] ) {
    int n = 0;
    for ( int i = 0; i < 9; i++ )
        n = n * 10 + a[ i ];
    return n;
}

int cantor( int a[] ) {
    int i, j, count, n = 0;
    for ( i = 0; i < 9; i++ ) {
        count = 0;
        for ( j = i + 1; j < 9; j++ ) {
            if ( a[ i ] > a[ j ] )
                count++;
        }
        n += count * factorial[ i ];
    }
    return n;
}


int main()
{
    int i;
    queue<int> q;
    int n, size, step;
    int temp[ 9 ], now[ 9 ];

    step = 0;
    q.push( 123456789 );
    while ( !q.empty() ) {
        size = q.size();
        while ( size-- ) {
            itoa( q.front(), now );

            rotate1( now, temp );
            n = cantor( temp );
            if ( tile[ n ] == 0 ) {
                tile[ n ] = step + 1;
                q.push( atoi( temp ) );
            }

            rotate2( now, temp );
            n = cantor( temp );
            if ( tile[ n ] == 0 ) {
                tile[ n ] = step + 1;
                q.push( atoi( temp ) );
            }

            rotate3( now, temp );
            n = cantor( temp );
            if ( tile[ n ] == 0 ) {
                tile[ n ] = step + 1;
                q.push( atoi( temp ) );
            }

            rotate4( now, temp );
            n = cantor( temp );
            if ( tile[ n ] == 0 ) {
                tile[ n ] = step + 1;
                q.push( atoi( temp ) );
            }

            q.pop();
        }
        step++;
    }
    tile[ 0 ] = 0;
    
    while ( scanf( "%d", &now[ 0 ] ) != EOF ) {
        for ( i = 1; i < 9; i++ )
            scanf( "%d", &now[ i ] );
        printf( "%d\n", tile[ cantor( now ) ] );
    }

    return 0;

}

void rotate1( int from[], int to[] ) {
    to[ 0 ] = from[ 1 ];
    to[ 1 ] = from[ 4 ];
    to[ 2 ] = from[ 2 ];
    to[ 3 ] = from[ 0 ];
    to[ 4 ] = from[ 3 ];
    to[ 5 ] = from[ 5 ];
    to[ 6 ] = from[ 6 ];
    to[ 7 ] = from[ 7 ];
    to[ 8 ] = from[ 8 ];
}

void rotate2( int from[], int to[] ) {
    to[ 0 ] = from[ 0 ];
    to[ 1 ] = from[ 2 ];
    to[ 2 ] = from[ 5 ];
    to[ 3 ] = from[ 3 ];
    to[ 4 ] = from[ 1 ];
    to[ 5 ] = from[ 4 ];
    to[ 6 ] = from[ 6 ];
    to[ 7 ] = from[ 7 ];
    to[ 8 ] = from[ 8 ];
}

void rotate3( int from[], int to[] ) {
    to[ 0 ] = from[ 0 ];
    to[ 1 ] = from[ 1 ];
    to[ 2 ] = from[ 2 ];
    to[ 3 ] = from[ 4 ];
    to[ 4 ] = from[ 7 ];
    to[ 5 ] = from[ 5 ];
    to[ 6 ] = from[ 3 ];
    to[ 7 ] = from[ 6 ];
    to[ 8 ] = from[ 8 ];
}

void rotate4( int from[], int to[] ) {
    to[ 0 ] = from[ 0 ];
    to[ 1 ] = from[ 1 ];
    to[ 2 ] = from[ 2 ];
    to[ 3 ] = from[ 3 ];
    to[ 4 ] = from[ 5 ];
    to[ 5 ] = from[ 8 ];
    to[ 6 ] = from[ 6 ];
    to[ 7 ] = from[ 4 ];
    to[ 8 ] = from[ 7 ];
}