URAL 1183. Brackets Sequence

Author: 2015-11-18

ACM

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1. 题目

http://acm.timus.ru/problem.aspx?space=1&num=1183

1183. Brackets Sequence

Time limit: 1.0 second
Memory limit: 64 MB

Let us define a regular brackets sequence in the following way:

  1. Empty sequence is a regular sequence.
  2. If S is a regular sequence, then (S) and [S] are both regular sequences.
  3. If A and B are regular sequences, then AB is a regular sequence.
For example, all of the following sequences of characters are regular brackets sequences:
(), [], (()), ([]), ()[], ()[()]
And all of the following character sequences are not:
(, [, ), )(, ([)], ([(]
Some sequence of characters ‘(‘, ‘)’, ‘[‘, and ‘]’ is given. You are to find the shortest possible regular brackets sequence, that contains the given character sequence as a subsequence. Here, a string a1a2…an is called a subsequence of the string b1b2…bm, if there exist such indices 1 ≤ i1 < i2 < … < in ≤ m, that aj=bij for all 1 ≤ j ≤ n.

Input

The input contains at most 100 brackets (characters ‘(‘, ‘)’, ‘[‘ and ‘]’) that are situated on a single line without any other characters among them.

Output

Write a single line that contains some regular brackets sequence that has the minimal possible length and contains the given sequence as a subsequence.

Sample

input output 
([(]
 
()[()]
Problem Author: Andrew Stankevich
Problem Source: 2001-2002 ACM Northeastern European Regional Programming Contest

2. 思路

给出一个由括号组成的序列，将其补充为平衡的括号序列，求最短的平衡括号序列。

区间DP。记输入为str[0 : L-1] ，使用int dp[i][j] 记录将str[i : j] 补充平衡所需的最少括号个数，使用string result[i][j] 记录将str[i : j] 补充平衡后的最短序列，空间消耗较大，但便于输出。对于str[i] 和str[j] ：

  1. 如果str[i] 和str[j] 相匹配（“()”或“[]”），则dp[i][j] = min(dp[i + 1][j – 1], dp[i][j]) ；
  2. 然后在str[i : j] 区间内进行DP，令dp[i][j] = min(dp[i][k] + dp[k + 1][j], dp[i][j])，k∈[i, j) 。

在更新dp[i][j] 的同时，要相应地更新result[i][j] 。注意测试例包含空行，scanf(“%s”, str) 无法读入空行会导致WA。

3. 代码

#include <iostream>
#include <string>
using namespace std;

const int MAX_BRACKETS_NUM = 101;
const int MAX = 1000000;

void solveP9f_BracketsSequence();

int main() {
    solveP9f_BracketsSequence();
    return 0;
}

int dp[MAX_BRACKETS_NUM][MAX_BRACKETS_NUM];
string result[MAX_BRACKETS_NUM][MAX_BRACKETS_NUM], str;
void solveP9f_BracketsSequence() {
    cin >> str;
    int len = str.length();

    for (int i = 0; i < len; i++) {
        for (int j = i; j < len; j++) {
            result[i][j] = "";
            dp[i][j] = MAX;
        }
    }

    for (int i = len - 1; i >= 0; i--) {
        for (int j = i; j < len; j++) {
            if (i == j) {
                dp[i][j] = 1;
                if (str[i] == '(' || str[i] == ')')
                    result[i][j] = "()";
                else 
                    result[i][j] = "[]";
            } else {
                if (str[i] == '(' && str[j] == ')' && (dp[i + 1][j - 1] < dp[i][j])) {
                    result[i][j] = "(" + result[i + 1][j - 1] + ")";
                    dp[i][j] = dp[i + 1][j - 1];
                } else if (str[i] == '[' && str[j] == ']' && (dp[i + 1][j - 1] < dp[i][j])) {
                    result[i][j] = "[" + result[i + 1][j - 1] + "]";
                    dp[i][j] = dp[i + 1][j - 1];
                }
                for (int k = i; k < j; k++) {
                    if (dp[i][j] > dp[i][k] + dp[k + 1][j]) {
                        dp[i][j] = dp[i][k] + dp[k + 1][j];
                        result[i][j] = result[i][k] + result[k + 1][j];
                    }
                }
            }
        }
    }
    
    cout << result[0][len - 1] << endl;
}