UVa 11552 - Fewest Flops

dp[m][i][j]代表第m组以字母(i +'a') 开头和以字母 ( j + 'a' )结尾的最小块数。kuai[m]中存储了第 m 组最少可以分为多少块，显然最少为所有相同字母放在一起时的块数，即这一组中不同字母的个数。

如果前一组的结尾与后一组的开头相同，则dp[m][i][j] = min( dp[m-1][st][ed] + kuai[m] - 1 );

若不同，则dp[m][i][j] = min(dp[m - 1][st][ed] + kuai[m] );

 

 1 #include <cstdio>

 2 #include <cstring>

 3 #include <cstdlib>

 4 #include <algorithm>

 5 

 6 using namespace std;

 7 

 8 const int MAXN = 1010;

 9 const int INF = 1 << 30;

10 

11 char str[MAXN];

12 int dp[MAXN][30][30];

13 int kuai[MAXN];       //记录每组之中最小的块数

14 bool vis[MAXN][30];   //记录每组出现了哪些字母

15 int K, n;

16 

17 void GetKuai()     //预处理得到每组内最小的块数

18 {

19     memset( vis, false, sizeof(vis) );

20     for ( int i = 0; i < n; ++i )

21     {

22         int st = K * i;

23         int ed = K * (i + 1);

24         for ( int j = st; j < ed; ++j )

25             vis[i][ str[j] - 'a' ] = true;

26 

27         kuai[i] = 0;

28         for ( int j = 0; j < 26; ++j )

29             if ( vis[i][j] ) ++kuai[i];

30     }

31     return;

32 }

33 

34 void init()     //初始化

35 {

36     for ( int i = 0; i < 26; ++i )

37         for ( int j = 0; j < 26; ++j )

38             if ( vis[0][i] && vis[0][j] )

39                 dp[0][i][j] = kuai[0];

40             else dp[0][i][j] = INF;

41     return;

42 }

43 

44 void solved()

45 {

46     init();

47 

48     for ( int m = 1; m < n; ++m )

49         for ( int i = 0; i < 26; ++i )

50         {

51             if ( vis[m][i] )

52             {

53                 for ( int j = 0; j < 26; ++j )

54                 {

55                     dp[m][i][j] = INF;

56                     if ( ( j == i && kuai[m] == 1 ) || ( j != i && vis[m][j] ) ) // 这里WA了两次，首位字母不能相同，除非这一组内的块数为1

57                     {

58                         for ( int st = 0; st < 26; ++st )

59                             if ( vis[m - 1][st] )

60                             {

61                                 for ( int ed = 0; ed < 26; ++ed )

62                                 {

63                                     if ( ( st == ed && kuai[m - 1] == 1 ) || ( ed != st && vis[m - 1][ed] ) )

64                                     {

65                                         if ( ed == i )

66                                             dp[m][i][j] = min( dp[m][i][j], dp[m - 1][st][ed] + kuai[m] - 1 );

67                                         else dp[m][i][j] = min( dp[m][i][j], dp[m - 1][st][ed] + kuai[m] );

68                                     }

69                                 }

70                             }

71                     }

72                 }

73             }

74         }

75 

76     int ans = INF;

77     for ( int i = 0; i < 26; ++i )

78         for ( int j = 0; j < 26; ++j )

79             if ( vis[n - 1][i] && vis[n - 1][j] )

80                 ans = min( ans, dp[n - 1][i][j] );

81 

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

83     return;

84 }

85 

86 int main()

87 {

88     int T;

89     scanf( "%d", &T );

90     while ( T-- )

91     {

92         scanf( "%d%s", &K, str );

93         n = strlen(str) / K;   //得到块数

94         GetKuai();

95         solved();

96     }

97     return 0;

98 }