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:

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 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_ij 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;
}