长度为N的任何括号序列的最小和

给定一个数字N表示由括号’(’，’)’组成的括号序列的长度。实际顺序事先未知。给定两个括号’(’和’)’的值(如果放在索引处) $i$在表达式中。

任务是使用上述信息找到长度为N的任何括号序列的最小和。

这里adj [i] [0]代表在第i个索引处分配给’)’括号的值，而adj [i] [1]代表在第i个索引处分配给’(’括号)的值。

限制条件：

• 应该有N / 2对用括号制成。也就是说，N / 2对’(’，’)’。
• 找到适当的括号表达式的最小和。
• 索引从0开始。

例子：

Input : N = 4 
        adj[N][2] ={{5000, 3000},
                    {6000, 2000}, 
                    {8000, 1000}, 
                    {9000, 6000}} 
Output : 19000
Assigning first index as '(' for proper 
bracket expression is (_ _ _  . 
Now all the possible bracket expressions are ()() and (()). 
where '(' denotes as adj[i][1] and ')' denotes as adj[i][0]. 
Hence, for ()() sum is 3000+6000+1000+9000=19000.
and (()), sum is 3000+2000+8000+9000=220000. 
Thus answer is 19000

Input : N = 4 
        adj[N][2] = {{435, 111},
                     {43, 33}, 
                     {1241, 1111}, 
                     {234, 22}}
Output : 1499

算法：

1. 方括号序列的第一个元素只能是’(’，因此adj [0] [1]的值仅在索引0处使用。
2. 如本文所述，使用dp调用函数以找到正确的括号表达式。
3. 将’(’表示为adj [i] [1]，将’)’表示为adj [i] [0]。
4. 找到所有可能的正确括号表达式的最小和。
5. 返回答案+ adj [0] [1]。

下面是上述方法的实现：

C++

// C++ program to find the Minimum sum possible
// of any bracket sequence of length N using
// the given values for brackets
  
#include 
using namespace std;
  
#define MAX_VAL 10000000
  
// DP array
int dp[100][100];
  
// Recusive function to check for
// correct bracket expression
int find(int index, int openbrk, int n, int adj[][2])
{
    /// Not a proper bracket expression
    if (openbrk < 0)
        return MAX_VAL;
  
    // If reaches at end
    if (index == n) {
  
        /// If proper bracket expression
        if (openbrk == 0) {
            return 0;
        }
        else // if not, return max
            return MAX_VAL;
    }
  
    // If alreaddy visited
    if (dp[index][openbrk] != -1)
        return dp[index][openbrk];
  
    // To find out minimum sum
    dp[index][openbrk] = min(adj[index][1] + find(index + 1,
                                                  openbrk + 1, n, adj),
                             adj[index][0] + find(index + 1,
                                                  openbrk - 1, n, adj));
  
    return dp[index][openbrk];
}
  
// Driver Code
int main()
{
    int n = 4;
    int adj[n][2] = { { 5000, 3000 },
                      { 6000, 2000 },
                      { 8000, 1000 },
                      { 9000, 6000 } };
  
    memset(dp, -1, sizeof(dp));
  
    cout << find(1, 1, n, adj) + adj[0][1] << endl;
  
    return 0;
}

Java

// Java program to find the Minimum sum possible 
// of any bracket sequence of length N using 
// the given values for brackets 
  
public class GFG {
  
    final static int MAX_VAL = 10000000 ;
  
    // DP array
    static int dp[][] = new int[100][100];
  
    // Recursive function to check for
    // correct bracket expression
    static int find(int index, int openbrk, int n, int adj[][])
    {
        /// Not a proper bracket expression
        if (openbrk < 0)
            return MAX_VAL;
  
        // If reaches at end
        if (index == n) {
  
            /// If proper bracket expression
            if (openbrk == 0) {
                return 0;
            }
            else // if not, return max
                return MAX_VAL;
        }
  
        // If alreaddy visited
        if (dp[index][openbrk] != -1)
            return dp[index][openbrk];
  
        // To find out minimum sum
        dp[index][openbrk] = Math.min(adj[index][1] + find(index + 1,
                                                      openbrk + 1, n, adj),
                                 adj[index][0] + find(index + 1,
                                                      openbrk - 1, n, adj));
  
        return dp[index][openbrk];
    }
  
  
// Driver code
    public static void main(String args[])
    {
            int n = 4;
            int adj[][] = { { 5000, 3000 },
                              { 6000, 2000 },
                              { 8000, 1000 },
                              { 9000, 6000 } };
  
            for (int i = 0; i < dp.length; i ++)
                for (int j = 0; j < dp.length; j++)
                    dp[i][j] = -1 ;
                  
  
            System.out.println(find(1, 1, n, adj) + adj[0][1]);
  
  
    }
    // This code is contributed by ANKITRAI1
}

Python3

# Python 3 program to find the Minimum sum 
# possible of any bracket sequence of length
# N using the given values for brackets
  
MAX_VAL = 10000000
  
# DP array
dp = [[-1 for i in range(100)]
          for i in range(100)]
  
# Recusive function to check for
# correct bracket expression
def find(index, openbrk, n, adj):
      
    # Not a proper bracket expression
    if (openbrk < 0):
        return MAX_VAL
  
    # If reaches at end
    if (index == n):
          
        # If proper bracket expression
        if (openbrk == 0):
            return 0
              
    # if not, return max
        else:
            return MAX_VAL
  
    # If alreaddy visited
    if (dp[index][openbrk] != -1):
        return dp[index][openbrk]
  
    # To find out minimum sum
    dp[index][openbrk] = min(adj[index][1] + find(index + 1, 
                                             openbrk + 1, n, adj), 
                             adj[index][0] + find(index + 1, 
                                             openbrk - 1, n, adj))
  
    return dp[index][openbrk]
  
# Driver Code
if __name__ == '__main__':
    n = 4;
    adj = [[5000, 3000],[6000, 2000],
           [8000, 1000],[9000, 6000]]
  
    print(find(1, 1, n, adj) + adj[0][1])
      
# This code is contributed by
# Sanjit_Prasad

C#

// C# program to find the Minimum sum possible 
// of any bracket sequence of length N using 
// the given values for brackets 
using System; 
    
class GFG 
{ 
    public static int MAX_VAL = 10000000;
      
    // DP array 
    public static int[,] dp = new int[100,100]; 
        
    // Recusive function to check for 
    // correct bracket expression 
    public static int find(int index, int openbrk, int n, int[,] adj) 
    { 
        /// Not a proper bracket expression 
        if (openbrk < 0) 
            return MAX_VAL; 
        
        // If reaches at end 
        if (index == n) { 
        
            /// If proper bracket expression 
            if (openbrk == 0) { 
                return 0; 
            } 
            else // if not, return max 
                return MAX_VAL; 
        } 
        
        // If alreaddy visited 
        if (dp[index,openbrk] != -1) 
            return dp[index,openbrk]; 
        
        // To find out minimum sum 
        dp[index,openbrk] = Math.Min(adj[index,1] + find(index + 1, 
                                                      openbrk + 1, n, adj), 
                                 adj[index,0] + find(index + 1, 
                                                      openbrk - 1, n, adj)); 
        
        return dp[index,openbrk]; 
    } 
        
    // Driver Code      
      
    static void Main() 
    { 
        int n = 4;
           
        int[,] adj = new int[,]{ 
                            { 5000, 3000 }, 
                            { 6000, 2000 }, 
                            { 8000, 1000 }, 
                            { 9000, 6000 } 
        }; 
        
        for(int i = 0; i < 100; i++)
            for(int j = 0; j < 100; j++)
                dp[i,j] = -1;
        
        Console.Write(find(1, 1, n, adj) + adj[0,1] + "\n"); 
    }
    //This code is contributed by DrRoot_
}

输出：

19000

时间复杂度：O(N ² )