TOPCODER SAM 686 div1 300

Posted 2020-07-14

// TOPCODER SAM 686 div1 300

Problem Statement

带有小中括号的括号序列，问可以去掉多少子串，使得剩下的非空串是合法的。

Constraints

字符串长度不超过 40。

Examples

// ans[i] = count(s[i])
string s[] = {"()[]", "())", "()()", "([)]", "())[]][]([]()]]()]]]"};
int ans[] = {3, 2, 4, 2, 3854};

Solution

寻找反面，等价于有多少非空的子串是合法的。这是一个区间 dp，不难转移。

Code

1. #include <string.h> 

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2.  

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3. #include <string> 

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4.  

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5. const int N = 123; 

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6.  

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7. class BracketSequenceDiv1 

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8. { 

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9. int n; 

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10. char str[N]; 

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11. long long dp[N][N]; 

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12.  

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13. public: 

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14. long long count(std::string std_str) 

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15. { 

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16. n = std_str.size(); 

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17. for (int i = 1; i <= n; ++i) { 

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18. str[i] = std_str[i-1]; 

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19. } 

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20. memset(dp, 0, sizeof(dp)); 

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21. return count(n); 

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22. } 

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23.  

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24. long long count(int n) 

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25. { 

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26. for (int d = 1; d <= n; ++d) { 

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27. for (int i = 1; i + d - 1 <= n; ++i) { 

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28. int j = i + d - 1; 

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29. dp[i][j] = dp[i][j-1]; 

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30. if (str[j] == ‘)‘) { 

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31. for (int k = i; k < j; ++k) { 

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32. // dp[i k-1] ( dp[k+1, j-1] ) 

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33. if (str[k] == ‘(‘) { 

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34. dp[i][j] += (dp[i][k-1] + 1) * (dp[k+1][j-1] + 1); 

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35. } 

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36. } 

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37. } 

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38. else if (str[j] == ‘]‘) { 

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39. for (int k = i; k < j; ++k) { 

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40. // dp[i, k-1] [ dp[k+1, j-1] ] 

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41. if (str[k] == ‘[‘) { 

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42. dp[i][j] += (dp[i][k-1] + 1) * (dp[k+1][j-1] + 1); 

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43. } 

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44. } 

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45. } 

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46. } 

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47. } 

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48. return dp[1][n]; 

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49. } 

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50. }; 

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