POJ 2955 Brackets

链接：http://poj.org/problem?id=2955

Brackets

Time Limit: 1000MS
Memory Limit: 65536KTotal 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 ofs. That is, you wish to find the largest m such that for indicesi₁, 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

66406

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;}