TOPCODER SAM 686 div1 300

// 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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3. #include <string> 
   

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5. const int N = 123; 
   

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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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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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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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