poj1141 2010.7.31

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最新推荐文章于 2020-12-23 03:30:20 发布

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本文探讨了一个关于括号序列的问题，通过详细解释其定义和规则，提出了一种寻找最短有效括号序列的方法，该序列包含给定序列作为子序列。通过实例演示并提供了C++代码实现，旨在简化复杂逻辑并提高效率。

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poj1141 2010.7.31

Brackets Sequence

Time Limit: 1000MS Memory Limit: 65536K

Total Submissions: 12467 Accepted: 3338 Special Judge

Description

Let us define a regular brackets sequencein 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, thenAB is a regular sequence.

For example, all of the following sequencesof characters are regular brackets sequences:

(), [], (()), ([]), ()[], ()[()]

And all of the following charactersequences are not:

(, [, ), )(, ([)], ([(]

Some sequence of characters '(', ')', '[',and ']' is given. You are to find the shortest possible regular bracketssequence, that contains the given character sequence as a subsequence. Here, astring a1 a2 ... an is called a subsequence of the string b1 b2 ... bm, ifthere exist such indices 1 = i1 < i2 < ... < in = m, that aj = bij forall 1 = j = n.

Input

The input file contains at most 100brackets (characters '(', ')', '[' and ']') that are situated on a single linewithout any other characters among them.

Output

Write to the output file a single line thatcontains some regular brackets sequence that has the minimal possible lengthand contains the given sequence as a subsequence.

Sample Input

([(]

Sample Output

()[()]

Source

Northeastern Europe 2001

详见lrj的黑书

不过，记录下来答案，用到c++里的<string>真的很不错~~~可以直接对字符串用+ 来衔接，不错~~~

ans[i][j]表示 i..j这断填好括号后的序列

WA 因：

1.dp方程敲成代码的时候，一激动敲错了

2.把"[]" 敲成了 ‘[]' ，酱紫就是字符，不是字符串了。。。

Source Code

Problem: 1141 User: creamxcream

Memory: 1020K Time: 391MS

Language: C++ Result: Accepted

Source Code

#include <string>
#include <cstdio>
#include <cstring>
#include <iostream>

#define INF ~0U>>1

using namespace std;

#define MAXN 106

string ans[MAXN][MAXN];
int dp[MAXN][MAXN];
char str[MAXN];
int n,flag=0;

void init()
{
	memset(dp,0,sizeof(dp));
	cin.getline(str+1,MAXN); 
	n=strlen(str+1);
    if (n==0)
	{
		flag=1;
		return;
	}
	for(int i=0;i<=n;i++)
	{
		dp[i][i]=1;
		if ((str[i]=='(')||(str[i]==')'))
			ans[i][i]="()";
		else
			ans[i][i]="[]";
	}
}

void dpit()
{
	for(int p=1;p<=n-1;p++)
		for(int i=1;i<=n-p;i++)
		{
			int j=i+p;
			dp[i][j]=INF;
			if (((str[i]=='(')&&(str[j]==')'))||
				((str[i]=='[')&&(str[j]==']')))
				if (dp[i+1][j-1]<dp[i][j])
				{
					dp[i][j]=dp[i+1][j-1];
					ans[i][j]=str[i]+ans[i+1][j-1]+str[j];
				}
			if ((str[i]=='(')||(str[i]=='['))
				if (dp[i+1][j]+1<dp[i][j])
				{
					dp[i][j]=dp[i+1][j]+1;
					if (str[i]=='(')
						ans[i][j]='('+ans[i+1][j]+')';
					else
						ans[i][j]='['+ans[i+1][j]+']';
				}
			if ((str[j]==')')||(str[j]==']'))
				if (dp[i][j-1]+1<dp[i][j])
				{
					dp[i][j]=dp[i][j-1]+1;
					if (str[j]==')')
						ans[i][j]='('+ans[i][j-1]+')';
					else
						ans[i][j]='['+ans[i][j-1]+']';
				}
			for (int k=i;k<=j-1;k++)
			{
				if (dp[i][k]+dp[k+1][j]<dp[i][j])
				{
					dp[i][j]=dp[i][k]+dp[k+1][j];
					ans[i][j]=ans[i][k]+ans[k+1][j];
				}
			}
		}
}


int main()
{
	init();
	if (flag==1)
	{
		printf("\n");
	}
	else
	{
		dpit();
		cout<<ans[1][n]<<endl;
	}
	return 0;
}