POJ 1141 Brackets Sequence

POJ 1141;链接：http://poj.org/problem?id=1141

题目：

Brackets Sequence

Time Limit: 1000MS		Memory Limit: 65536K
Total Submissions: 23638		Accepted: 6665		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

()[()]

题目分析：

用dp得到记忆表path[][],用递归求出最短的合法序列。

注意点：建议用gets（）来读入字符串，因为测试数据中有空行，如果用scanf（）或者cin读不到空格（太坑，因为这个wa了很多次）；

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

const int MAXN = 110;
char str[MAXN];
int dp[MAXN][MAXN], path[MAXN][MAXN];

void oprint(int i, int j)
{
	if(i > j) return ;
	if(i == j)
	{
		if('[' == str[i] || ']' == str[i])
			cout << "[]";
		else
			cout << "()";
	}
	else 
	{
		if(-1 == path[i][j])
		{
			cout << str[i];
			oprint(i + 1, j - 1);
			cout << str[j];
		}
		else
		{
			oprint(i, path[i][j]);
			oprint(path[i][j]+1, j);
		}
	}
}

int main()
{
	std::ios::sync_with_stdio(false);
	
	while(gets(str))
	{
		int n = strlen(str);
		
		if(0 == n) 
		{
			cout << endl;
			continue;
		}

		memset(dp, 0, sizeof(dp));
		memset(path, 0, sizeof(path));
		
		for(int i = 0; i < n; i++)
			dp[i][i] = 1;
			
		for(int r = 1; r < n; r++)
		{
			for(int i = 0; i < n - r; i++)
			{
				
				int j = i + r;
				dp[i][j] = 0x7fffffff;
				
				if( ('(' == str[i] && ')' == str[j]) || ('[' == str[i] && ']' == str[j]) )
				{
					dp[i][j]   = dp[i + 1][j - 1];
					path[i][j] = -1;
				//	continue;//不能有continue
				}
				
				for(int k = i; k < j; k++)
				{
					if(dp[i][j] > dp[i][k] + dp[k+1][j])
					{
						dp[i][j]   = dp[i][k] + dp[k+1][j];
						path[i][j] = k;
					}
				}
				
			}
		}
		
		oprint(0, n - 1);
		cout << endl;
	}

	return 0;
}