poj1141（区间DP）

地址：http://poj.org/problem?id=1141

Brackets Sequence

Time Limit: 1000MS		Memory Limit: 65536K
Total Submissions: 23828		Accepted: 6714		Special Judge

Description

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.
Sample Input

([(]

Sample Output

()[()]

题意：问最少添加几个括号使每个括号有可以匹配的另一半，输出修改后的串。

思路：区间DP，和另一道有点像，不过这里要输出修改的串。输出串用递归就行了。

代码：

#include<iostream>
#include<queue>
#include<algorithm>
#include<cstdio>
#include<cstring>
using namespace std;
#define LL __int64
#define Mod 0xfffffff
char bk[110];
struct node
{
int ans,k;
} dp[110][110];
void pr(int a,int b)  //递归输出
{
if(a>b) return ;
else if(dp[a][b].ans==0)
{
for(int i=a; i<=b; i++)
{
if(bk[i]=='('||bk[i]==')') printf("()");
if(bk[i]=='['||bk[i]==']') printf("[]");
}
}
else if(dp[a][b].ans==b-a+1)
{
for(int i=a; i<=b; i++)
printf("%c",bk[i]);
}
else if(dp[a][b].k==b)
{
printf("%c",bk[a]);
pr(a+1,dp[a][b].k-1);
printf("%c",bk[dp[a][b].k]);
}
else
{
pr(a,dp[a][b].k);
pr(dp[a][b].k+1,b);
}
}
int main()
{
gets(bk);  //注意不要用while(scanf!=EOF)，用scanf输入的记得在%s后加\n
int len=strlen(bk);
for(int i=0; i<len; i++)
for(int j=0; j<len; j++)
dp[i][j].ans=0;
for(int l=1; l<len; l++)
{
for(int i=0; i<len-l; i++)
{
int j=i+l;
for(int k=i+1; k<=j; k++)
{
if((bk[i]=='('&&bk[k]==')')||(bk[i]=='['&&bk[k]==']'))
{
if(dp[i][j].ans<dp[i+1][k-1].ans+dp[k+1][j].ans+2)
{
dp[i][j].ans=dp[i+1][k-1].ans+dp[k+1][j].ans+2;
dp[i][j].k=k;
}
}
else
{
if(dp[i][j].ans<dp[i][k-1].ans+dp[k][j].ans)
{
dp[i][j].ans=dp[i][k-1].ans+dp[k][j].ans;
dp[i][j].k=k-1;
}
}
}
}
}
pr(0,len-1);
puts("");
return 0;
}