hdu4681String

http://acm.hdu.edu.cn/showproblem.php?pid=4681

枚举A串和B串包含C串的区间  枚举区间端点算左右两端最长公共子序

 1 #include <iostream>

 2 #include<cstdio>

 3 #include<cstring>

 4 #include<stdlib.h>

 5 #include<algorithm>

 6 using namespace std;

 7 #define N 1010

 8 char s1[N],s2[N],s3[N];

 9 int dp1[N][N],dp2[N][N];

10 struct node

11 {

12     int l,r;

13 }p1[N],p2[N];

14 int main()

15 {

16     int i,j,t,k1,k2,k3,kk=0;

17     scanf("%d",&t);

18     while(t--)

19     {

20         kk++;

21         memset(dp1,0,sizeof(dp1));

22         memset(dp2,0,sizeof(dp2));

23         scanf("%s%s%s",s1,s2,s3);

24         k1 = strlen(s1);

25         k2 = strlen(s2);

26         k3 = strlen(s3);

27         for(i = 1 ; i <= k1 ; i++)

28             for(j = 1; j <= k2 ;j++)

29             {

30                 if(s1[i-1]==s2[j-1])

31                 dp1[i][j] = dp1[i-1][j-1]+1;

32                 else

33                 dp1[i][j] = max(dp1[i-1][j],dp1[i][j-1]);

34             }

35         for(i = k1 ; i>= 1 ; i--)

36             for(j = k2 ; j >= 1 ; j--)

37             {

38                 if(s1[i-1]==s2[j-1])

39                 dp2[i][j] = dp2[i+1][j+1]+1;

40                 else

41                 dp2[i][j] = max(dp2[i+1][j],dp2[i][j+1]);

42             }

43         int g = 0;j=0;

44         for(i = 0 ; i < k1 ; i++)

45         if(s1[i]==s3[j])

46         {

47             j++;int a=i;

48             if(j<k3)

49             {

50                 for(a = i+1 ; a < k1 ; a++)

51                 {

52                     if(s1[a]==s3[j])

53                     j++;

54                     if(j==k3)

55                     break;

56                 }

57             }

58             if(j==k3)

59             {

60                 g++;

61                 p1[g].l = i;

62                 p1[g].r = a+1;

63             }

64             j = 0;

65         }

66         int o=0;j=0;

67         for(i = 0 ; i < k2 ; i++)

68         if(s2[i]==s3[j])

69         {

70             j++;int a=i;

71             if(j<k3)

72             {

73                 for(a = i+1 ; a < k2 ; a++)

74                 {

75                     if(s2[a]==s3[j])

76                     j++;

77                     if(j==k3)

78                     break;

79                 }

80             }

81             if(j==k3)

82             {

83                 o++;

84                 p2[o].l = i;

85                 p2[o].r = a+1;

86             }

87             j = 0;

88         }

89         int maxz=0;

90         for(i = 1; i <= g ; i++)

91             for(j = 1 ; j <= o ; j++)

92             {

93                 int x  = dp1[p1[i].l][p2[j].l],x2 = dp2[p1[i].r+1][p2[j].r+1];

94                 maxz = max(maxz,x+x2);

95             }

96         printf("Case #%d: %d\n",kk,maxz+k3);

97     }

98     return 0;

99 }

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