UVA1626 / ZOJ1463 Brackets sequence 区间DP

简单区间DP (有空串... ...)

Brackets sequence

Time Limit: 4500MS

Memory Limit: Unknown

64bit IO Format: %lld & %llu

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Description





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 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.

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 

()[()]

Source

Root :: AOAPC II: Beginning Algorithm Contests (Second Edition) (Rujia Liu) :: Chapter 9. Dynamic Programming :: Examples

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/* ***********************************************
Author :CKboss
Created Time :2015年02月11日 星期三 16时15分08秒
File Name :ZOJ1463.cpp
************************************************ */

#include <iostream>
#include <cstdio>
#include <cstring>
#include <algorithm>
#include <string>
#include <cmath>
#include <cstdlib>
#include <vector>
#include <queue>
#include <set>
#include <map>

using namespace std;

const int maxn=300;
const int INF=0x3f3f3f3f;

char str[maxn];
int n;
int dp[maxn][maxn];

bool match(int a,int b)
{
if((str[a]=='('&&str[b]==')')||(str[a]=='['&&str[b]==']'))
return true;
return false;
}

void PRINT(int l,int r)
{
if(l==r)
{
if(str[l]=='('||str[l]==')')
{
putchar('('); putchar(')');
}
if(str[l]=='['||str[l]==']')
{
putchar('['); putchar(']');
}
return ;
}
else if(l>r) return ;
int pos=-1;
int temp=INF;
if(match(l,r)) temp=dp[l+1][r-1];
for(int i=l;i+1<=r;i++)
if(dp[l][i]+dp[i+1][r]<temp)
{
pos=i; temp=dp[l][i]+dp[i+1][r];
}
if(pos==-1)
{
putchar(str[l]); PRINT(l+1,r-1); putchar(str[r]);
}
else
{
PRINT(l,pos); PRINT(pos+1,r);
}
}

int main()
{
//freopen("in.txt","r",stdin);
//freopen("out.txt","w",stdout);

int T_T,flag=0;
scanf("%d",&T_T);
getchar();
while(T_T--)
{
gets(str);
memset(str,0,sizeof(str));
gets(str);
n=strlen(str);
if(n==0)
{
if(flag++) putchar(10);
putchar(10);
continue;
}

/// DP
memset(dp,63,sizeof(dp));
for(int i=0;i<n;i++) dp[i][i]=1;
for(int i=0;i<n;i++)
for(int j=0;j<n;j++)
if(i>j) dp[i][j]=0;
for(int len=2;len<=n;len++)
{
for(int i=0;i+len-1<n;i++)
{
/// i...j
int j=i+len-1;
if(match(i,j)) dp[i][j]=min(dp[i][j],dp[i+1][j-1]);
for(int k=i;k+1<=j;k++)
dp[i][j]=min(dp[i][j],dp[i][k]+dp[k+1][j]);
}
}
if(flag++) putchar(10);
PRINT(0,n-1);
putchar(10);
}

return 0;
}