UVa 1626 Brackets sequence

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 

()[()]

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

// 原始表达式
char str[105];
//string str;

// record[i][j]代表从i到j形成表达式需要添加的最小长度
int record[110][110];

// next[i][j]代表从i到j形成表达式时i配对的对应括号的位置
// 如果next[i][j]为-1代表自行添加，(,[,),]添加成(),[].
int next[110][110];

int get_len(int i, int j);

void print_result(int begin, int end);

int len;

/*
FILE* fp;
FILE* fp2;
*/

void readline(char* S) {
  fgets(S, 105, stdin);
}

int main()
{
	int n;
//	fp = fopen("1.txt", "r");
//	fp2 = fopen("2.txt", "wr");
//	scanf("%d", &n);
/*	cin >> n;
	char ch;
	while((ch=getchar()) != '\n')

*/		;
//	fscanf(fp, "%d", &n);
	readline(str);
  	sscanf(str, "%d", &n);
  	readline(str);
	int count = 0;
	while(count < n)
	{
		memset(record, -1, sizeof(record));
//		memset(next, -1, sizeof(next));
//		memset(str, 0, sizeof(str));
//		scanf("%s", str);
//		getline(cin, str);
		readline(str);
		len = strlen(str) - 1;
//		cout << "str:" << str << endl;
//		getline(cin, str);
//		cout << "str:" << str << endl;	
//		fscanf(fp, "%s", str);
//		len = strlen(str);
//		len = str.length();	
		// 计算结果
//		printf("result: %d\n", get_len(0, len-1));	
//		get_len(0, len-1);		
/*	
		for(int i = 0; i < len; i++)
		{
			for(int j = i; j < len; j++)
				printf("result(%d,%d): %d\n", i, j, record[i][j]);
		}	
*/		// 打印结果
		if(count > 0)
		{
			printf("\n");
//			fprintf(fp2, "\n");
		}
/*		for(int i = 0; i < len; i++)
			printf("%c: %d\n", str[i], next[i]);
*/	
		print_result(0, len-1);	
		printf("\n");
		readline(str);
		count++;

	}
/*	fclose(fp);
      	fclose(fp2);
*/	return 0;
}

// 计算结果
// 从i到j形成表达式需要添加的最小长度
int get_len(int i, int j)
{

//	printf("here: record[%d][%d]\n", i, j);
	if(i > j)
		return 0;

	if(record[i][j] != -1)
		return record[i][j];

//	printf("here: record[%d][%d]\n", i, j);
	if(i == j)
	{
//		next[i][i] = -1;
		record[i][j] = 1;
//		printf("here: record[%d][%d]: %d\n", i, j, record[i][j]);
//		printf("(%d,%d): %d\n", i, j, record[i][j]);
		return record[i][j];	
	}

	record[i][j] = len;
/*	char ch = '#';
	if(str[i] == '(')
		ch = ')';
	else if(str[i] == '[')
		ch = ']';

	int result = 1 + get_len(i+1, j);
        if(result < record[i][j])
        {
                record[i][j] = result;
                next[i][j] = -1;
        }
	for(int end = i+1; end <= j; end++)
	{
		if(str[end] == ch)
		{
			int result;
			result = 0;
			if(end - i > 1)
				result = get_len(i+1, end-1);
			if(j - end >= 1)
				result += get_len(end+1, j);
			if(result < record[i][j])
			{
				record[i][j] = result;
				next[i][j] = end;
			}	 	 
		}			
	}		 
*/
//	int result = (1<<30);
	if(str[i] == '(' && str[j] == ')')
	{
/*		if(i+1 < j)	
			result = get_len(i+1, j-1);
		else
			result = 0;
*/		record[i][j] = min(record[i][j], get_len(i+1,j-1));
	}
	else if(str[i] == '[' && str[j] == ']')
	{
/*		if(i+1 < j)
			result = get_len(i+1, j-1);
		else
			result = 0;
*/		record[i][j] = min(record[i][j], get_len(i+1, j-1));
	}
/*
	if(result < record[i][j])
	{
		record[i][j] = result;
//		next[i][j] = j;
	}	
*/
	for(int begin = i; begin <= j-1; begin++)
	{
		int result = get_len(i, begin) + get_len(begin+1, j);

		if(result < record[i][j])
			record[i][j] = result;		
	}

	return record[i][j];
}


// 打印结果
void print_result(int begin, int end)
{
	if(begin > end)
		return;
//	if(next[begin][end] == -1)
	if(begin == end)
	{
		if(str[begin] == '(' || str[begin] == ')')
		{
			printf("()");
//			fprintf(fp2, "()");
		}
		else
		{ 
			printf("[]");
//			fprintf(fp2, "[]");
		}
		return;
/*		if(begin+1 <= end)
			print_result(begin+1, end);
*/	}	
/*	else if(((begin+1==end)||(record[begin][end] == record[begin+1][end-1]))&& 
		((str[begin]=='('&&str[end]==')')||(str[begin]=='['&&str[end]==']')))
*/
	int ans = get_len(begin, end);
/*	else if(record[begin][end]==record[begin+1][end-1] &&
		((str[begin]=='('&&str[end]==')')||(str[begin]=='['&&str[end]==']')))
*/
	if(ans == get_len(begin+1, end-1) && ((str[begin]=='('&&str[end]==')')||(str[begin]=='['&&str[end]==']')))
	{
		printf("%c", str[begin]);
//		fprintf(fp2, "%c", str[begin]);
/*
		int mid = next[begin][end];
		if(begin+1 <= mid-1)
			print_result(begin+1, mid-1);
		printf("%c", str[mid]);
//		fprintf(fp2, "%c", str[mid]);
		if(mid+1 <= end)
			print_result(mid+1, end);
*/	
//		if(begin+1 <= end-1)
			print_result(begin+1, end-1);
		printf("%c", str[end]);		
		return;
	}
/*	else
	{
*/		for(int i = begin; i <= end-1; i++)
		{
			if(ans == get_len(begin,i)+get_len(i+1,end))
			{
				print_result(begin, i);
				print_result(i+1, end);
				return;
			}
		}
/*	}

	if(begin == 0 && end == len-1)
	{
		printf("\n");
//		fprintf(fp2, "\n");
	}
*/

}

这次代码写的很烂。思路错了，应该依照表达式构造方法来想。

分两种可能，如果S -> (S') 或者 [S']，检查S'

还有S -> AB，检查AB.