POJ 2488 A Knight's Journey

DFS，要求输出字典序最小的，注意扩展方向。

# include <cstdio>
# include <cstring>

# define N 26 + 5

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

int p, q, cnt;
bool finished;
char solu[N][2];
char vis[N][N];

void dfs(int x, int y)
{
    solu[cnt][0] = x, solu[cnt][1] = y;
    if (cnt == p*q)
    {
        finished = true;
        return ;
    }
    else
    {
        for (int i = 0; i < 8; ++i)
        {
            int nx = x + dx[i];
            int ny = y + dy[i];
            if (1<=nx&&nx<=p && 1<=ny&&ny<=q && !vis[nx][ny])
            {
                vis[nx][ny] = 1, ++cnt;
                dfs(nx, ny);
                if (finished) return;
                vis[nx][ny] = 0, --cnt;
            }
        }
    }

}

void solve(void)
{
    finished = false;

    for (int i = 1; i <= p; ++i)
                memset(vis[i]+1, 0, sizeof(int)*q);

    vis[1][1] = 1;
    cnt = 1;
    dfs(1, 1);
    if (finished)
    {
        printf("A1");
        for (int i = 2; i <= p*q; ++i)
            printf("%c%c", solu[i][1]-1+'A', solu[i][0]-1+'1');
        putchar('\n');
    }
    else
        puts("impossible");
}

int main()
{
    int T;

//    freopen("PKU2488.in", "r", stdin);

    scanf("%d", &T);
    for (int i = 1; i <= T; ++i)
    {
        if (i > 1) putchar('\n');
        printf("Scenario #%d:\n", i);
        scanf("%d%d", &p, &q);
        solve();
    }

    return 0;
}