poj 1141 Brackets Sequence(区间dp)

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最新推荐文章于 2021-09-13 20:44:00 发布

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分类专栏： DP 动态规划文章标签：区间dp

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Brackets Sequence

Time Limit: 1000MS		Memory Limit: 65536K
Total Submissions: 26424		Accepted: 7451		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

()[()]

Source

Northeastern Europe 2001

区间dp

#include<iostream>
#include<cstdio>
#include<cstring>
#include<algorithm>
using namespace std;

#define N 305
#define INF 0x3f3f3f3f

int dp[N][N],vis[N][N];
char a[N];

inline bool judge(int i,int j)
{
	if(a[i]=='('&&a[j]==')'||a[i]=='['&&a[j]==']')
		return true;
	return false;
}

void print(int i,int j)
{
	if(i>j) return ;
	if(i==j)
	{
		if(a[i]=='('||a[i]==')')
			printf("()");
		else
			printf("[]");
		return ;
	}

	if(vis[i][j]==-1)
	{
		printf("%c",a[i]);
		print(i+1,j-1);
		printf("%c",a[j]);
		return ;
	}

    print(i,vis[i][j]);
    print(vis[i][j]+1,j);

}

int main()
{
	int i,j;
	while(gets(a))
	{
		int len=strlen(a);
		if(len==0)
		{
			printf("\n");
			continue;
		}
		memset(dp,0,sizeof(dp));
		memset(vis,-1,sizeof(vis));

		for(i=0;i<len;i++)
			dp[i][i]=1;

		for(i=len-1;i>=0;i--)
			for(j=i+1;j<len;j++)
		{
			dp[i][j]=dp[i+1][j]+1;         //刚开始是自己匹配，即加一个
			vis[i][j]=i;

            if(judge(i,j))                       //如果首尾匹配,这个特殊处理
			{
				if(dp[i][j]>dp[i+1][j-1])
				{
					dp[i][j]=dp[i+1][j-1];
					vis[i][j]=-1;
				}
			}
			for(int k=i+1;k<j;k++)               //i~j，之间有匹配的
				if(judge(i,k))
			{
				if(dp[i][j]>dp[i][k]+dp[k+1][j])
				{
					dp[i][j]=dp[i][k]+dp[k+1][j];
					vis[i][j]=k;
				}
			}
		}

		print(0,len-1);
		printf("\n");
	}
	return 0;
}