UVA - 1626 Brackets sequence ( dp )

原题：

Let us defie 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 fid the shortest possible
regular brackets sequence, that contains the given character sequence as a subsequence. Here, a string
a 1a
2. . . a
n
is called a subsequence of the stringb
1b
2. . . b
m, if there exist such indices1
≤i
1 < i2
<
. . . < in≤
m, thataj=
bij
for all1
≤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.
The input fie 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 fie 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：

 
     给一行由'('')''['']'组成的序列，求最少添加几个字符可以使整个序列全部括号匹配。

思路：

 
     取dp[i][j]表示字串s[i][j]至少需要添加多少个括号，当s[i]和s[j]匹配时，dp[i] [j]=min{ dp[i] [j] , dp[i+1] [j-1] }。dp[i] [j]=min{ dp[i] [j] , dp[i] [k]+dp[k+1] [j] | i<=k<j }。题目还有两个坑点，一个是可能输出空串，并且输入空串时还应对应输出一个空串；另一个是每个相邻样例间要有一个输出一个空行，并且最后一个样例后不输出空行。导致WA了好几遍……

#include <iostream>
#include <iomanip>
#include <algorithm>
#include <cstdio>
#include <cstring>
#include <queue>
#include <deque>
#include <string>
#include <cmath>
#include <vector>
#include <utility>
#include <set>
#include <map>
#include <climits>
//#pragma comment(linker, "/STACK:1024000000,1024000000")
#define INF 2147483647
using namespace std;
typedef long long ll;
int t;
string s;
int n,dp[105][105];
bool match(string ss,int a,int b)
{
if(ss[a]=='('&&ss[b]==')')
return true;
else if(ss[a]=='['&&ss[b]==']')
return true;
else
return false ;
}
void print(int i,int j)
{
if(i>j)
return ;
if(i==j)
{
if(s[i]=='('||s[i]==')')
printf("()");
else
printf("[]");
return ;
}
int ans=dp[i][j];
if(match(s,i,j)&&ans==dp[i+1][j-1])
{
cout<<s[i];
print(i+1,j-1);
cout<<s[j];
return ;
}
for(int k=i;k<j;k++)
if(ans==dp[i][k]+dp[k+1][j])
{
print(i,k);
print(k+1,j);
return ;
}

}
int main()
{
scanf("%d",&t);
getchar();
while(t--)
{
getchar();
s.clear();
//cin>>s;
getline(cin,s);
n=s.size();
memset(dp,0,sizeof(dp));
for(int i=0; i<n; i++)
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,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)//最后一个样例不输出空行
printf("\n");
}
return 0;
}