TOPCODER SRM 686 div2 1000

// TOPCODER SRM 686 div2 1000

Problem Statement

给出一个至多长 100 的字符串，仅包含 ( 和 )，问其中有多少个不重复的，合法的括号子序列。

子序列可以不连续；合法即括号序列的合法；答案模 1,000,000,007。

Examples

"(())("
Returns: 2

Correct non-empty bracket subsequences are "()" and "(())".

"())"
Returns: 1

Only "()" is suitable.

")((("
Returns: 0

There are no non-empty correct bracket subsequences.

"()()()()()()()()(())))(()()()()))())"
Returns: 364675

"()()()()()()()()()()()()()()()()()()()()()()()()()()()()()()()()()()()()()()()()()()()()()()()()()()"
Returns: 122826009

Solution

f[i][j][c] 表示对于前 i 个字符组成的串，左括号比右括号多 j 个，并且以 c 为结尾的子序列的个数。注意这里的状态存的不是所有的子序列，在转移的时候为了让括号合法，仅转移 j >= 0 的那些状态，这个时候 f[i][j][c] 的值比所有的子序列要少，而 f[i][0][1] 就是前 i 个字符组成的串中，合法的，不重复子序列数目。

这道问题跟 CODEFROCES 645E 的模型是同一个。

http://www.cnblogs.com/gu-castle/p/5535969.html

Code

1. public class BracketSequenceDiv2 {

   ---

2. static final int MOD = (int) 1e9 + 7; 
   

   ---

3. ---

4. int modAdd(int a, int b) { 
   

   ---

5. int sum = a + b; 
   

   ---

6. if (sum >= MOD) { 
   

   ---

7. sum -= MOD; 
   

   ---

8. } 
   

   ---

9. return sum; 
   

   ---

10. } 
    

    ---

11. ---

12. public int count(String str) { 
    

    ---

13. int n = str.length(); 
    

    ---

14. int[][][] dp = new int[n + 1][n + 1][2]; 
    

    ---

15. ---

16. for (int i = 1; i <= n; ++i) { 
    

    ---

17. char ch = str.charAt(i - 1); 
    

    ---

18. if (ch == ‘(‘) { 
    

    ---

19. dp[i][1][0] = modAdd(dp[i - 1][0][0], dp[i - 1][0][1]); 
    

    ---

20. dp[i][1][0] = modAdd(dp[i][1][0], 1); 
    

    ---

21. for (int j = 2; j <= i; ++j) { 
    

    ---

22. dp[i][j][0] = modAdd(dp[i - 1][j - 1][0], 
    

    ---

23. dp[i - 1][j - 1][1]); 
    

    ---

24. } 
    

    ---

25. for (int j = 0; j <= i; ++j) { 
    

    ---

26. dp[i][j][1] = dp[i - 1][j][1]; 
    

    ---

27. } 
    

    ---

28. } 
    

    ---

29. else { // ch == ‘)‘ 
    

    ---

30. for (int j = 0; j < i; ++j) { 
    

    ---

31. dp[i][j][1] = modAdd(dp[i - 1][j + 1][0], 
    

    ---

32. dp[i - 1][j + 1][1]); 
    

    ---

33. } 
    

    ---

34. for (int j = 0; j <= i; ++j) { 
    

    ---

35. dp[i][j][0] = dp[i - 1][j][0]; 
    

    ---

36. } 
    

    ---

37. } 
    

    ---

38. } 
    

    ---

39. return dp[n][0][1]; 
    

    ---

40. } 
    

    ---

41. } 
    

    ---