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 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.
This problem contains multiple test cases!
The first line of a multiple input is an integer N, then a blank line followed by N input blocks. Each input block is in the format indicated in the problem description. There is a blank
line between input blocks.
The output format consists of N output blocks. There is a blank line between output blocks.
Sample Input
1
([(]
Sample Output
()[()]
关键在于输入与输出格式。神坑。
区间dp,dp[i][j]表示
区间 i 到j之间的匹配数，区间两端的 字符能否够刚好匹配,若能够匹配 状态转移就多了一个 dp[i][j] = max(dp[i][k]+dp[k+1][j],dp[i+1][j-1]+1),若不能匹配就是dp[i][j]
= max(dp[i][j],dp[i][k]+dp[k+1][j]);
若是两端能够匹配的，并且两端匹配了导致的dp值最大那么就标记一下。mark[i][j]
= -1,否则 就mark[i][j] = k，这样把全部区间都dp一遍，回头再用DFS寻找。若是两端匹配导致值最大的 那么就直接输出这个字符标记一下，继续往更小的区间去搜索，否则 就分开两个区间搜索 [i,k] [k+1,j]
#include <stdio.h>
#include <string.h>
#include <algorithm>
using namespace std;
#define up(i,x,y) for(i=x;i<=y;i++)
#define down(i,x,y) for(i=x;i>=y;i--)
#define mem(a,b) memset(a,b,sizeof(a))
#define w(a) while(a)
char str[105];
int t,len,dp[105][105],mark[105][105],pos[105];
void dfs(int i,int j)
{
if(mark[i][j]==-1)
{
pos[i]=pos[j]=1;
dfs(i+1,j-1);
}
else if(mark[i][j]>=0)
{
dfs(i,mark[i][j]);
dfs(mark[i][j]+1,j);
}
return;
}
int main()
{
int l,i,j,k;
scanf("%d%*c%*c",&t);
while(t--)
{
gets(str);
len=strlen(str);
if(!len)
{
printf("\n");
if(t)
printf("\n");
continue;
}
up(i,0,len-1)
up(j,0,len-1)
{
mark[i][j]=-2;
dp[i][j]=0;
}
mem(pos,0);
i=j=l=0;
w(l<len)
{
if(i==j)
{
i++,j++;
if(j==len)
i=0,l++,j=l;
continue;
}
if((str[i]=='('&&str[j]==')')||(str[i]=='['&&str[j]==']'))
{
up(k,i,j-1)
{
if(dp[i][j]<dp[i][k]+dp[k+1][j])
{
mark[i][j]=k;
dp[i][j]=dp[i][k]+dp[k+1][j];
}
}
if(dp[i][j]<dp[i+1][j-1]+1)
{
mark[i][j]=-1;
dp[i][j]=dp[i+1][j-1]+1;
}
}
else
{
up(k,i,j-1)
{
if(dp[i][j]<dp[i][k]+dp[k+1][j])
{
mark[i][j]=k;
dp[i][j]=dp[i][k]+dp[k+1][j];
}
}
}
i++,j++;
if(j==len)
{
l++;
i=0;
j=l;
}
}
dfs(0,len-1);
up(i,0,len-1)
{
if(pos[i]==1)
printf("%c",str[i]);
else if(str[i]=='('||str[i]==')')
printf("()");
else
printf("[]");
}
printf("\n");
if(t)
{
printf("\n");
getchar();
}
} return 0;
}
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