Let us define a regular brackets sequence in the following way:
- Empty sequence is a regular sequence.
- If S is a regular sequence, then (S) and [S] are both regular sequences.
- 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 a1 a2 … an is called a subsequence of the string b1 b2 … bm, if there exist such indices 1 = i1 < i2 < … < in = m, that aj = bij for all 1 = j = n.
Input
The input file contains at most 100 brackets (characters ‘(’, ‘)’, ‘[’ and ‘]’) that are situated on a single line without any other characters among them.
Output
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
([(]
Sample Output
()[()]
分析：
题意：
给出一个字符串（可以是空串），让所有的括号都能正确匹配（本串内的字符无法匹配的通过插入相应的字符来解决），求正确匹配的字符串最短的字符串！
分析：
对于这道题，我们初始所有的字符都不能与字符串本身的字符正确匹配，这是最坏的情况。我们用一个数组dp[i][j]表示字符串中从字符str[i]~str[j]的字符串中最少要插入多少个字符：
(1)如果str[i]与str[j]配对，则：dp[i][j]=dp[i+1][j-1];
这也可能不是要插入的最少字符，例如：[][]
他的str[1]就与str[4]匹配，但是dp[2][3]的值为2，通过观察，这个字符串本身不用插入任何字符，所以我们还要通过以下操作：
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];
}
}
这样我们就知道了字符串最少要插入的字符数，但是这道题求的是插入后的字符串，而非要插入的字符的最少数！那怎么办呢？可能大家都想到了，那就是做标记，哪个位置需要插入字符就在哪个位置做标记！我们在定义一个数组：mark[i][j]。
（1）如果mark[i][j]的值为0，那么表示字符str[i]和str[j]是匹配的，那么我们就先输出str[i],再对str[i+1]~str[j-1]的字符做进一步处理；
（2）如果mark[i][j]的值不为0，那么他记录的就是与字符str[i]匹配的字符的位置，那么我们就从mark[i][j]这里划分开来，将两段字符串分开处理！
代码：
#include<iostream>
#include<cstring>
#include<cstdio>
#define INF 0x3f3f3f3f
#define N 105
using namespace std;
int dp[N][N],mark[N][N];
char str[N];
bool cmp(char a,char b)//判断是否匹配
{
if((a=='('&&b==')')||(a=='['&&b==']'))
return true;
return false;
}
void Print(int l,int r)//输出递归函数
{
if(l>r)
return;
if(l==r)
{
if(str[l-1]=='['||str[l-1]==']')
printf("[]");
else
printf("()");
return;
}
if(!mark[l][r])
{
printf("%c",str[l-1]);
Print(l+1,r-1);
printf("%c",str[r-1]);
}
else
{
Print(l,mark[l][r]);
Print(mark[l][r]+1,r);
}
}
int main()
{
while(gets(str)!=NULL)
{
int len=strlen(str);
memset(dp,0,sizeof(dp));
for(int i=1;i<=len;i++)
{
dp[i][i]=1;
}
for(int len1=2;len1<=len;len1++)
{
for(int i=1;i+len1-1<=len;i++)
{
int j=len1+i-1;
dp[i][j]=INF;
if(cmp(str[i-1],str[j-1]))
{
dp[i][j]=dp[i+1][j-1];
mark[i][j]=0;
}
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];
mark[i][j]=k;
}
}
}
}
Print(1,len);
printf("\n");
}
return 0;
}
/*([][[])*/