1427. SMS(DP)

１４２７

题意不太好理解　其它没什么　细心啊　细心　一个０写成了１　ＷＡ半天

以每个字符是以第一种方式还是第二种方式来Ｄ

 1 #include <iostream>

 2 #include<cstdio>

 3 #include<algorithm>

 4 #include<stdlib.h>

 5 #include<cstring>

 6 #include<vector>

 7 using namespace std;

 8 #define N 100010

 9 char s[N];

10 int dp[N][2],o[N][2];

11 int judge(char x)

12 {

13     if(x==' ')

14     return 1;

15     if(x>='a'&&x<='z')

16     return 1;

17     if(x>='A'&&x<='Z')

18     return 1;

19     return 0;

20 }

21 int main()

22 {

23     int n,m,i,k=0;

24     scanf("%d%d%*c",&n,&m);

25     gets(s);

26     k = strlen(s);

27     dp[0][1] = 1;

28     o[0][1] = 1;

29     if(judge(s[0]))

30     {

31         dp[0][0] = 1;

32         o[0][0] = 1;

33     }

34     else

35     {

36         dp[0][0] = -1;

37         o[0][0] = 0;

38     }

39     for(i = 1 ; i < k ; i++)

40     {

41         if(judge(s[i]))

42         {

43             if(o[i-1][0]>=m)

44             {

45                 dp[i][0] = min(dp[i-1][1]+1,dp[i-1][0]+1);

46                 o[i][0] = 1;

47             }

48             else

49             {

50                 if(o[i-1][0]&&dp[i-1][0]<dp[i-1][1]+1)

51                 {

52                     dp[i][0] = dp[i-1][0];

53                     o[i][0] = o[i-1][0]+1;

54                 }

55                 else

56                 {

57                     dp[i][0] = dp[i-1][1]+1;

58                     o[i][0] = 1;

59                 }

60             }

61         }

62         else

63             o[i][0] = 0;

64         if(o[i-1][1]>=n)

65         {

66             if(o[i-1][0])

67             dp[i][1] = min(dp[i-1][1]+1,dp[i-1][0]+1);

68             else

69             dp[i][1] = dp[i-1][1]+1;

70             o[i][1] = 1;

71         }

72         else

73         {

74             if(!o[i-1][0]||dp[i-1][1]<dp[i-1][0]+1)

75             {

76                 dp[i][1] = dp[i-1][1];

77                 o[i][1] = o[i-1][1]+1;

78             }

79             else

80             {

81                 dp[i][1] = dp[i-1][0]+1;

82                 o[i][1] = 1;

83             }

84         }

85 

86     }

87     if(o[k-1][0]!=0)

88     printf("%d\n",min(dp[k-1][0],dp[k-1][1]));

89     else

90     printf("%d\n",dp[k-1][1]);

91     return 0;

92 }

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