题目
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 begins with a single positive integer on a line by itself indicating the number of the cases
following, each of them as described below. This line is followed by a blank line, and there is also a
blank line between two consecutive inputs.//好坑，之前读的中文题意，这部分没有！！！
The input file contains at most 100 brackets (characters ‘(’, ‘)’, ‘[’ and ‘]’) that are situated on a
single line without any other characters among them.
Output
For each test case, the output must follow the description below. The outputs of two consecutive cases
will be separated by a blank line.
Write to the output file 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
1
([(]
Sample Output
()[()]
题意
定义如下正规括号序列（字符串）：
- 空序列是正规括号序列。
- 如果S是正规括号序列，那么(S)和[S]也是正规括号序列。
- 如果A和B都是正规括号序列，那么AB也是正规括号序列。
例如，下面的字符串都是正规括号序列：()，[]，(())，([])，()[]，()[()]，而如下字符串 则不是正规括号序列：(，[，]，)(，([()。
输入一个长度不超过100的，由“(”、“)”、“[”、“]”构成的序列，添加尽量少的括号，得 到一个规则序列。有多解输一个序列。
注意事项
坑点来了
- 输入时要多输一个字符（只能用getchar();因为那个字符好像并不是'\n',不能用scanf("\n");）
- 输入字符组s时要输两遍因为很可能有空串
- 因为空串所以不能用scanf("%s",s)输入，只能用getchar,fgets,getline//算法书上这道题的最后有一排小字就写的这个，可惜我眼瞎（
作者心黑），我之前用gets结果CE编译错误……，可能是它不得UVA的宠爱╭(╯^╰)╮，之后我别无他法，只能痛下杀手（痛快下手），用了fgets，希望它在天之灵能明白我的苦衷（都是UVA的错）o(￣ヘ￣o＃)哼哼~。 - 输出的时候我放松了警惕，以为套路适可而止，可是UVA一点也不知道分寸。它要输换行（这里还没毛病）可是它要两个，两个，两个！！！并且，最后一组数据只输出一个换行。
（你可能觉得这些都是细节，可是，只要你一个没对，它就WA,WA,WA！！！）
思路
（终于到这了，真心不容易）
状态定义：
dp[i][j]:表示从s[i~j]至少需要添加几个括号
状态转移方程式：
if(match(s[i],s[j]))
dp[i][j]=min(dp[i][j],dp[i+1][j-1]);
for(int k=i;k<j;k++)//以k做断点（必须枚举判最优，不管是否匹配）
dp[i][j]=min(dp[i][j],dp[i][k]+dp[k+1][j]);
代码
#include<cstdio>
#include<algorithm>
#include<cstring>
using namespace std;
#define INF 0x3f
int t,n;
char s[102];
int dp[102][102];
bool match(char a,char b)
{return (a=='('&&b==')')||(a=='['&&b==']');
}
void print(int i,int j)
{if(i>j)return;if(i==j){if(s[i]=='('||s[j]==')')printf("()");else printf("[]");return;}int a=dp[i][j];if(match(s[i],s[j])&&a==dp[i+1][j-1]){printf("%c",s[i]);print(i+1,j-1);printf("%c",s[j]);return;}for(int k=i;k<j;k++)if(a==dp[i][k]+dp[k+1][j]){print(i,k);print(k+1,j);return;}
}
int main()
{scanf("%d",&t);while(t--){fgets(s,102,stdin);fgets(s,102,stdin);n=strlen(s)-1;memset(dp,0,sizeof(dp));for(int i=0;i<n;i++)dp[i+1][i]=0,dp[i][i]=1;for(int i=n-2;i>=0;i--)for(int j=i+1;j<n;j++){dp[i][j]=n;if(match(s[i],s[j]))dp[i][j]=min(dp[i][j],dp[i+1][j-1]);for(int k=i;k<j;k++)dp[i][j]=min(dp[i][j],dp[i][k]+dp[k+1][j]);}print(0,n-1);printf("\n");if(t)printf("\n");}
}