poj 1141【dp--记录路径】

 

Brackets Sequence

Time Limit: 1000MS		Memory Limit: 65536K
Total Submissions: 16818		Accepted: 4571		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

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

#include 
using namespace std;

char s[102];
int dp[110][110], path[110][110];
const int inf = 0xffffff;

void output(int i, int j)
{
	if (i > j)
	return;
	if (i == j)
	{
		if (s[i] == '(' || s[j] == ')') printf("()");
		else printf("[]");
	}
	else
	if (path[i][j] == -1)
	{
		printf("%c",s[i]);
		output(i+1,j-1);
		printf("%c",s[j]);
	}

	else
	{
		output(i,path[i][j]);
		output(path[i][j]+1,j);
	}
}

int main()
{
	scanf("%s",s);
	memset(dp,0,sizeof(dp));
	int n = strlen(s);
	for (int i = 0; i < n; ++i)
	dp[i][i] = 1;

	for (int p = 1; p < n; ++p)
	for (int i = 0; i < n-p; ++i)
	{
		int j = i+p;
		dp[i][j] = inf;
		if ((s[i] == '(' && s[j] == ')') || (s[i] == '[' && s[j] == ']'))
		if(dp[i][j]>dp[i+1][j-1])
                        dp[i][j]=dp[i+1][j-1],path[i][j]=-1;
		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;
			}
		}
	}
	output(0,n-1);
	printf("\n");
 
	return 0;
}