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...a_n is called a subsequence of the string b₁b₂...b_m, if there exist such indices 1 ≤ i₁ < i₂ < ... < i_n ≤ m, that a_j=b_{i_j} 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; }/// DPmemset(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...jint 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;
}