UVa 1291 - Dance Dance Revolution

dp[k][i][j]表示第k步左脚在位置 i，右脚在位置 j 状态时的最小能量消耗.

dp[k][i][j] = min( dp[ k - 1 ][x][j] + cost[x][ step[k] ], dp[ k - 1 ][i][x] + cost[x][ step[k] ] );

这题很坑爹的没有数据范围，数组我随便开的，居然1Y了……

 1 #include <cstdio>

 2 #include <cstring>

 3 #include <cstdlib>

 4 #include <algorithm>

 5 

 6 using namespace std;

 7 

 8 const int MAXN = 10010;

 9 const int INF = 1 << 30;

10 

11 int step[MAXN];

12 int dp[MAXN][6][6];

13 int cost[6][6];

14 int N;

15 

16 void init()

17 {

18     for ( int i = 0; i < 5; ++i ) cost[i][i] = 1;

19     for ( int i = 1; i < 5; ++i ) cost[0][i] = 2;

20     for ( int i = 1; i < 5; ++i )

21     {

22         cost[i][i + 1] = 3;

23         cost[i + 1][i] = 3;

24     }

25     cost[1][4] = 3;

26     cost[4][1] = 3;

27     cost[1][3] = 4;

28     cost[3][1] = 4;

29     cost[2][4] = 4;

30     cost[4][2] = 4;

31 

32     return;

33 }

34 

35 void DP()

36 {

37     dp[0][0][0] = 0;

38     for ( int k = 1; k < N; ++k )

39     {

40         for ( int i = 0; i <= 4; ++i )

41         {

42             int j = step[k];

43             for ( int x = 0; x <= 4; ++x )

44             {

45                 dp[k][j][i] = min( dp[k][j][i], dp[k - 1][x][i] + cost[x][j] );

46                 dp[k][i][j] = min( dp[k][i][j], dp[k - 1][i][x] + cost[x][j] );

47             }

48         }

49     }

50 

51     int ans = INF;

52     for ( int i = 0; i <= 4; ++i )

53     for ( int j = 0; j <= 4; ++j )

54         ans = min( ans, dp[N - 1][i][j] );

55 

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

57 

58     return;

59 }

60 

61 int main()

62 {

63     init();

64     int a;

65     while( scanf( "%d", &a ), a )

66     {

67         step[1] = a;

68         N = 1;

69         while ( scanf( "%d", &step[ ++N ] ), step[N] );

70 

71         for ( int i = 0; i <= N; ++i )

72             for ( int j = 0; j <= 4; ++j )

73                 for ( int k = 0; k <= 4; ++k )

74                     dp[i][j][k] = INF;

75 

76         DP();

77     }

78     return 0;

79 }