POJ 2488 A Knight's Journey

题目链接：http://poj.org/problem?id=2488

分析：简单DFS，注意字典序最小

#include <cstdio>
#include <cstring>

const int dx[] = { -1, 1, -2, 2, -2, 2, -1, 1 };
const int dy[] = { -2, -2, -1, -1, 1, 1, 2, 2 };
const int MAXN = 30;

struct point
{
    int x, y;
};

bool vis[MAXN][MAXN];
point path[MAXN];
int p, q;
int all;

bool check( int& x, int& y )
{
    return ( x >= 0 && x < p && y >= 0 && y < q );
}

bool DFS( int x, int y, int num )
{
//    printf("x=%d y=%d num=%d\n", x, y, num);

    path[num - 1].x = x;
    path[num - 1].y = y;

    if ( num == all )
        return true;

    vis[x][y] = true;

    for ( int i = 0; i < 8; i++ )
    {
        int xx = x + dx[i];
        int yy = y + dy[i];
        if ( check( xx, yy ) && !vis[xx][yy] )
        {
            if ( DFS( xx, yy, num + 1 ) ) return true;
        }
    }

    vis[x][y] = false;

    return false;
}

int main()
{
    int T, tt = 0;
//    freopen( "s.txt", "w", stdout );
    scanf( "%d", &T );
    while ( T-- )
    {
        scanf( "%d%d", &p, &q );
        all = p * q;
        bool flag = false;
        for ( int j = 0; j < q; j++ )
        {
            for ( int i = 0; i < p; i++ )
            {
                memset( vis, false, sizeof(vis) );
                if ( DFS(i, j, 1) )
                {
                    flag = true;
                    break;
                }
            }
            if ( flag ) break;
        }

        printf( "Scenario #%d:\n", ++tt );

        if ( flag )
        {
            for ( int i = 0; i < all; i++ )
                printf("%c%d", path[i].y + 'A', path[i].x + 1 );
            putchar('\n');
        }
        else puts("impossible");
        putchar('\n');
    }
    return 0;
}