# 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

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) 。

# 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[len - 1] << endl;
}