题解 | 括号区间匹配

#include <bits/stdc++.h>
using namespace std;

int dp[105][105];

bool match(char left, char right) {
    return (left == '(' && right == ')') || (left == '[' && right == ']');
}

int main() {
    string s;
    cin >> s;
    int n = s.length();
    
    // 初始化
    for(int i = 0; i < n; i++) {
        for(int j = 0; j < n; j++) {
            dp[i][j] = n;  // 初始化为最大可能值
        }
        dp[i][i] = 1;  // 单个括号需要插入一个配对括号
    }
    
    // 处理长度为2的情况
    for(int i = 0; i < n-1; i++) {
        if(match(s[i], s[i+1])) {
            dp[i][i+1] = 0;  // 如果匹配，不需要插入
        } else {
            dp[i][i+1] = 2;  // 不匹配，需要插入两个括号
        }
    }
    
    // 区间dp
    for(int len = 3; len <= n; len++) {
        for(int i = 0; i + len - 1 < n; i++) {
            int j = i + len - 1;
            
            // 如果两端匹配
            if(match(s[i], s[j])) {
                dp[i][j] = dp[i+1][j-1];
            } else {
                // 在左端或右端插入一个括号
                dp[i][j] = min(dp[i+1][j] + 1, dp[i][j-1] + 1);
            }
            
            // 枚举分割点
            for(int k = i; k < j; k++) {
                dp[i][j] = min(dp[i][j], dp[i][k] + dp[k+1][j]);
            }
        }
    }
    
    cout << dp[0][n-1] << endl;
    return 0;
}

import java.util.*;

public class Main {
    static int[][] dp = new int[105][105];
    
    static boolean match(char left, char right) {
        return (left == '(' && right == ')') || (left == '[' && right == ']');
    }
    
    public static void main(String[] args) {
        Scanner sc = new Scanner(System.in);
        String s = sc.next();
        int n = s.length();
        
        // 初始化
        for(int i = 0; i < n; i++) {
            Arrays.fill(dp[i], n);  // 初始化为最大可能值
            dp[i][i] = 1;  // 单个括号需要插入一个配对括号
        }
        
        // 处理长度为2的情况
        for(int i = 0; i < n-1; i++) {
            if(match(s.charAt(i), s.charAt(i+1))) {
                dp[i][i+1] = 0;
            } else {
                dp[i][i+1] = 2;
            }
        }
        
        // 区间dp
        for(int len = 3; len <= n; len++) {
            for(int i = 0; i + len - 1 < n; i++) {
                int j = i + len - 1;
                
                // 如果两端匹配
                if(match(s.charAt(i), s.charAt(j))) {
                    dp[i][j] = dp[i+1][j-1];
                } else {
                    // 在左端或右端插入一个括号
                    dp[i][j] = Math.min(dp[i+1][j] + 1, dp[i][j-1] + 1);
                }
                
                // 枚举分割点
                for(int k = i; k < j; k++) {
                    dp[i][j] = Math.min(dp[i][j], dp[i][k] + dp[k+1][j]);
                }
            }
        }
        
        System.out.println(dp[0][n-1]);
        sc.close();
    }
}

def match(left: str, right: str) -> bool:
    return (left == '(' and right == ')') or (left == '[' and right == ']')

def solve(s: str) -> int:
    n = len(s)
    dp = [[n] * n for _ in range(n)]  # 初始化为最大可能值
    
    # 初始化单个括号的情况
    for i in range(n):
        dp[i][i] = 1
    
    # 处理长度为2的情况
    for i in range(n-1):
        if match(s[i], s[i+1]):
            dp[i][i+1] = 0
        else:
            dp[i][i+1] = 2
    
    # 区间dp
    for length in range(3, n+1):
        for i in range(n-length+1):
            j = i + length - 1
            
            # 如果两端匹配
            if match(s[i], s[j]):
                dp[i][j] = dp[i+1][j-1]
            else:
                # 在左端或右端插入一个括号
                dp[i][j] = min(dp[i+1][j] + 1, dp[i][j-1] + 1)
            
            # 枚举分割点
            for k in range(i, j):
                dp[i][j] = min(dp[i][j], dp[i][k] + dp[k+1][j])
    
    return dp[0][n-1]

def main():
    s = input().strip()
    print(solve(s))

if __name__ == "__main__":
    main()