POJ 2955 Brackets

链接：http://poj.org/problem?id=2955

Brackets

Time Limit: 1000MS		Memory Limit: 65536K
Total Submissions: 4378		Accepted: 2331

Description

We give the following inductive definition of a “regular brackets” sequence:

• the empty sequence is a regular brackets sequence,
• if s is a regular brackets sequence, then (s) and [s] are regular brackets sequences, and
• if a and b are regular brackets sequences, then ab is a regular brackets sequence.
• no other sequence is a regular brackets sequence

For instance, all of the following character sequences are regular brackets sequences:

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

while the following character sequences are not:

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

Given a brackets sequence of characters a₁a₂ … a_n, your goal is to find the length of the longest regular brackets sequence that is a subsequence of s. That is, you wish to find the largest m such that for indices i₁, i₂, …, i_m where 1 ≤ i₁ < i₂ < … < i_m ≤ n, a_i1a_i2 … a_im is a regular brackets sequence.

Given the initial sequence ([([]])], the longest regular brackets subsequence is [([])].

Input

The input test file will contain multiple test cases. Each input test case consists of a single line containing only the characters (, ), [, and ]; each input test will have length between 1 and 100, inclusive. The end-of-file is marked by a line containing the word “end” and should not be processed.

Output

For each input case, the program should print the length of the longest possible regular brackets subsequence on a single line.

Sample Input

((()))
()()()
([]])
)[)(
([][][)
end

Sample Output

6
6
4
0
6

Source

Stanford Local 2004

大意——给你一个长度不超过100的括号序列，求最长合法括号子序列的长度。合法的序列判定为：1.空的序列是合法的；2.如果一个序列s是合法的，那么(s)和[s]都是合法的；3.如果两个序列a和b分别都是合法的，那么ab也是合法的。例如：(), [], (()), ()[], ()[()]都是合法的，而(, ], )(, ([)], ([(]都不是合法的。

思路——典型区间DP模型。分析问题可以发现：如果找到一对匹配的括号[xxx]oooo，就把区间分成两部分，一部分是xxx，另一部分是oooo。设dp[i][j]表示区间[i,j]之间的最长合法括号子序列长度，那么当i<j时，如果区间[i+1,j]内没有与i匹配的括号，则dp[i][j]=dp[i+1][j]；如果存在一个匹配的k，那么dp[i][j] = max{dp[i+1][j],dp[i+1][k-1]+dp[k+1][j]+
2(i+1<k<j&&i和k是一对匹配的括号)}。因此，我们把i从串末枚举到串首即可，最后dp[0][len-1]即为答案，len表示串长度。

复杂度分析——时间复杂度:O(len^3),空间复杂度:O(len^2)

附上AC代码：

#include <iostream>
#include <cstdio>
#include <string>
#include <cmath>
#include <iomanip>
#include <ctime>
#include <climits>
#include <cstdlib>
#include <cstring>
#include <algorithm>
#include <queue>
#include <vector>
#include <set>
#include <map>
//#pragma comment(linker, "/STACK:102400000, 102400000")
using namespace std;
typedef unsigned int li;
typedef long long ll;
typedef unsigned long long ull;
typedef long double ld;
const double pi = acos(-1.0);
const double e = exp(1.0);
const double eps = 1e-8;
const int maxlen = 105;
char str[maxlen];
int dp[maxlen][maxlen];

bool match(char a, char b); // 是否满足括号匹配

int main()
{
	ios::sync_with_stdio(false);
	while (scanf("%s", str) && strcmp("end", str))
	{
		memset(dp, 0, sizeof(dp)); // 初始化
		int len = strlen(str);
		for (int i=len-1; i>=0; i--)
		{ // 从后面不断扩大区间
			for (int j=i+1; j<len; j++)
			{ // dp[i][j]表示区间[i,j]之间的最长合法括号子序列长度
				dp[i][j] = dp[i+1][j]; // 区间[i+1,j]没有与i匹配的括号
				for (int k=i+1; k<=j; k++) // 寻找到匹配，一分为二(匹配内外)
					if (match(str[i], str[k])) // 比较大小，取大的
						dp[i][j] = max(dp[i][j], dp[i+1][k-1]+dp[k+1][j]+2);
			}
		}
		printf("%d\n", dp[0][len-1]);
	}
	return 0;
}

bool match(char a, char b)
{
	if ((a=='('&&b==')') || (a=='['&&b==']'))
		return 1;
	return 0;
}