poj_1141

括号匹配，结果为匹配后的串

 

一、记忆化搜索(search with finger)：

从[0, strlen(s)-1] 开始搜索，缩短区间，期间记录搜索过的值，防止重复

cut[l][r]代表区间[l, r]中需要从哪里分割，为输出

串长不超过100 复杂度可以接受

 

 1 // Brackets sequence

 2 // Search with fingers

 3 

 4 #include <iostream>

 5 #include <string>

 6 #include <algorithm>

 7 #include <cstring>

 8 #include <cstdio>

 9 

10 #define MAXN 100

11 

12 int n, dp[MAXN+10][MAXN+10], cut[MAXN+10][MAXN+10];

13 char s[111];

14 

15 int SearchWithFinger(int l, int r)

16 {

17     if(l > r)

18         return dp[l][r] = 0;

19     if(l == r)

20         return dp[l][r] = 1;

21     if(dp[l][r]!=-1)

22         return dp[l][r];

23 

24     int num = 1<<30;

25     if( (s[l]=='('&&s[r]==')')||(s[l]=='['&&s[r]==']') )

26         num = std::min(num, SearchWithFinger(l+1, r-1));

27     else if(s[l]=='('||s[l]=='[')

28         num = std::min(num, SearchWithFinger(l+1, r)+1);

29     else if(s[r]==')'||s[r]==']')

30         num = std::min(num, SearchWithFinger(l, r-1)+1);

31 

32     for(int m = l; m < r; ++m){

33         int temp = SearchWithFinger(l, m) + SearchWithFinger(m+1, r);

34         if(num >= temp){

35             num = temp;

36             cut[l][r] = m;

37         }

38     }

39 

40     return dp[l][r] = num;

41 

42 }

43 

44 void PrintOut(int l, int r)

45 {

46     if(l > r)

47         return;

48     if(l == r){

49         if(s[l]=='('||s[l]==')')

50             printf("()");

51         if(s[l]=='['||s[l]==']')

52             printf("[]");

53     }

54     else{

55         if(cut[l][r] >= 0){

56             PrintOut(l, cut[l][r]);

57             PrintOut(cut[l][r]+1, r);

58         }

59         else{

60             if(s[l]=='(' ){

61                 printf("(");

62                 PrintOut(l+1, r-1);

63                 printf(")");

64             }

65             if(s[l]=='['){

66                 printf("[");

67                 PrintOut(l+1, r-1);

68                 printf("]");

69             }

70         }

71     }

72 

73 }

74 

75 int main(int argc, char const *argv[])

76 {

77     // freopen("in", "r", stdin);

78     // freopen("out", "w", stdout);

79     scanf("%d", &n);

80     getchar();

81     while(n--){

82 

83         getchar();

84         gets(s);

85     

86 

87         memset(dp, -1, sizeof(dp));

88         memset(cut, -1, sizeof(cut));

89 

90         SearchWithFinger(0, strlen(s)-1);

91 

92         PrintOut(0, strlen(s)-1);

93 

94         printf("\n");

95         if(n)

96             printf("\n");

97     }

98     return 0;

99 }

 

二、区间DP

dp[l][r]代表从区间[l, r]的串需要添加至少多少个括号才能匹配

枚举区间长度len ： 1 to strlen(s)-1

dp[i][i] = 0 任何一个单独的括号， 都需要添加至少一个括号才能匹配

cut[l][r]意义同上

 1 // Brackets sequence

 2 // dynamic programming

 3 

 4 #include <cstdio>

 5 #include <cstring>

 6 #include <string>

 7 #include <algorithm>

 8 #include <iostream>

 9 

10 #define MAXN 100

11 

12 int n, dp[MAXN+5][MAXN+5], cut[MAXN+5][MAXN+5];

13 char s[MAXN+10];

14 

15 void PrintOut(int l, int r)

16 {

17     if(l > r)

18         return ;

19     if(l == r){

20         if(s[l]=='('||s[l]==')')

21             printf("()");

22         if(s[l]=='['||s[l]==']')

23             printf("[]");

24     }

25     else{

26         if(cut[l][r] >= 0){

27             PrintOut(l, cut[l][r]);

28             PrintOut(cut[l][r]+1, r);

29         }

30         else{

31             if(s[l]=='(' ){

32                 printf("(");

33                 PrintOut(l+1, r-1);

34                 printf(")");

35             }

36             if(s[l]=='['){

37                 printf("[");

38                 PrintOut(l+1, r-1);

39                 printf("]");

40             }

41         }

42     }

43 

44 }

45 

46 int main(int argc, char const *argv[])

47 {

48     // freopen("in", "r", stdin);

49 

50     while(gets(s)){

51 

52         memset(dp, 0, sizeof(dp));

53         memset(cut, -1, sizeof(cut));

54 

55         for(int i = 0; i < strlen(s); ++i)

56             dp[i][i] = 1;

57 

58         for(int len = 1; len < strlen(s); ++len){

59             for(int l = 0; l+len < strlen(s); ++l){

60                 int r = l+len;

61                 dp[l][r] = 1<<30;

62                 if( (s[l]=='('&&s[r]==')')||(s[l]=='['&&s[r]==']') )

63                     dp[l][r] = std::min(dp[l][r], dp[l+1][r-1]);

64                 else if(s[l]=='('||s[l]=='[')

65                     dp[l][r] = std::min(dp[l][r], dp[l+1][r]+1);

66                 else if(s[r]==')'||s[r]==']')

67                     dp[l][r] = std::min(dp[l][r], dp[l][r-1]+1);

68                 for(int m = l; m < r; ++m){

69                     if(dp[l][r] >= dp[l][m] + dp[m+1][r]){

70                         dp[l][r] = dp[l][m] + dp[m+1][r];

71                         cut[l][r] = m;

72                     }

73                 }

74             }

75         }

76         // printf("%d\n", dp[0][strlen(s)-1]);

77         PrintOut(0, strlen(s)-1);

78         printf("\n");

79     }

80     return 0;

81 }