[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])

 1 /*

 2     UVA 10051 - Tower of Cubes

 3 */

 4 # include <cstdio>

 5 # include <cstring>

 6 

 7 # define N 500 + 5

 8 # define F 6

 9 

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

11 

12 int n;

13 int a[N][F];

14 int f[N][F];

15 

16 int max(int x, int y)

17 {

18     return x>y ? x:y;

19 }

20 

21 int dp(int i, int p)

22 {

23     int &ans = f[i][p];

24     if (ans > 0) return ans;

25     ans = 1;

26     for (int j = 1; j < i; ++j)

27     for (int q = 0; q < F; ++q)

28         if (a[j][q] == a[i][p^1])

29             ans = max(ans, dp(j, q)+1);

30     return ans;

31 }

32 

33 void print_ans(int i, int p)

34 {

35     for (int j = 1; j < i; ++j)

36     for (int q = 0; q < F; ++q)

37     {

38         if (a[j][q] == a[i][p^1] && f[j][q]+1 == f[i][p])

39         {

40             print_ans(j, q);

41             printf("%d %s\n", j, s[q^1]);

42             j = i+1;

43             break;

44         }

45     }

46 }

47 

48 void init(void)

49 {

50     for (int i = 1; i <= n; ++i)

51     for (int j = 0; j < F; ++j)

52         scanf("%d", &a[i][j]);

53     memset(f, -1, sizeof(f[0])*(n+1));

54 }

55 

56 void solve(int icase)

57 {

58     if (icase != 1) putchar('\n');

59     int ans = 0, ti, tj;

60     printf("Case #%d\n", icase);

61     for (int i = 1; i <= n; ++i)

62     for (int j = 0; j < 6; ++j)

63     {

64         // printf("%d\n", dp(i, j));

65         if (ans < dp(i, j))

66         {

67             ans = f[i][j];

68             ti = i;

69             tj = j;

70         }

71     }

72     printf("%d\n", ans);

73     print_ans(ti, tj);

74     printf("%d %s\n", ti, s[tj^1]);

75 }

76 

77 int main()

78 {    

79     int icase = 0;

80     while (scanf("%d", &n), n)

81     {

82         init();

83         solve(++icase);

84     }

85     

86     return 0;

87 }