// 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

#include <string.h>

#include <string>
 
const int N = 123;
 
class BracketSequenceDiv1
{
  int n;
  char str[N];
  long long dp[N][N];
 
public:
  long long count(std::string std_str)
  {
    n = std_str.size();
    for (int i = 1; i <= n; ++i) {
      str[i] = std_str[i-1];
    }
    memset(dp, 0, sizeof(dp));
    return count(n);
  }
 
  long long count(int n)
  {
    for (int d = 1; d <= n; ++d) {
      for (int i = 1; i + d - 1 <= n; ++i) {
        int j = i + d - 1;
        dp[i][j] = dp[i][j-1];
        if (str[j] == ')') {
          for (int k = i; k < j; ++k) {
            // dp[i k-1] ( dp[k+1, j-1] )
            if (str[k] == '(') {
              dp[i][j] += (dp[i][k-1] + 1) * (dp[k+1][j-1] + 1);
            }
          }
        }
        else if (str[j] == ']') {
          for (int k = i; k < j; ++k) {
            // dp[i, k-1] [ dp[k+1, j-1] ]
            if (str[k] == '[') {
              dp[i][j] += (dp[i][k-1] + 1) * (dp[k+1][j-1] + 1);
            }
          }
        }
      }
    }
    return dp[1][n];
  }
};