[简单DP] UVA 10051 Tower of Cubes

DAG上的DP，六个方向，打印不要求字典序，相反方向可以通过异或得到（定义0-5代表题目所说的五个方向）；

f[i, p] = 1;

for (k = 1:i-1)

　　f[i, p] = max(f[i, p], f[k, q]+1) if (color[i][p^1] == color[k][q])

/*
    UVA 10051 - Tower of Cubes
*/
# include <cstdio>
# include <cstring>

# define N 500 + 5
# define F 6

const char s[F][10] = {"front", "back", "left", "right", "top", "bottom"};

int n;
int a[N][F];
int f[N][F];

int max(int x, int y)
{
    return x>y ? x:y;
}

int dp(int i, int p)
{
    int &ans = f[i][p];
    if (ans > 0) return ans;
    ans = 1;
    for (int j = 1; j < i; ++j)
    for (int q = 0; q < F; ++q)
        if (a[j][q] == a[i][p^1])
            ans = max(ans, dp(j, q)+1);
    return ans;
}

void print_ans(int i, int p)
{
    for (int j = 1; j < i; ++j)
    for (int q = 0; q < F; ++q)
    {
        if (a[j][q] == a[i][p^1] && f[j][q]+1 == f[i][p])
        {
            print_ans(j, q);
            printf("%d %s\n", j, s[q^1]);
            j = i+1;
            break;
        }
    }
}

void init(void)
{
    for (int i = 1; i <= n; ++i)
    for (int j = 0; j < F; ++j)
        scanf("%d", &a[i][j]);
    memset(f, -1, sizeof(f[0])*(n+1));
}

void solve(int icase)
{
    if (icase != 1) putchar('\n');
    int ans = 0, ti, tj;
    printf("Case #%d\n", icase);
    for (int i = 1; i <= n; ++i)
    for (int j = 0; j < 6; ++j)
    {
        // printf("%d\n", dp(i, j));
        if (ans < dp(i, j))
        {
            ans = f[i][j];
            ti = i;
            tj = j;
        }
    }
    printf("%d\n", ans);
    print_ans(ti, tj);
    printf("%d %s\n", ti, s[tj^1]);
}

int main()
{    
    int icase = 0;
    while (scanf("%d", &n), n)
    {
        init();
        solve(++icase);
    }
    
    return 0;
}