POJ 2955 Brackets 【区间dp】【水题】

Brackets

Time Limit: 1000MS Memory Limit: 65536KTotal Submissions: 8027 Accepted: 4264

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

66406

Source

Stanford Local 2004

dp[i][j]表示区间i~j中匹配的个数

#include<iostream>#include<algorithm>#include<cmath>#include<queue>#include<cstring>#include<iomanip>#define INF 0x3f3f3f3f#define ms(a,b) memset(a,b,sizeof(a))using namespace std;const int maxn = 105;char s[maxn];int dp[maxn][maxn];bool check(int i, int j){if (s[i] == '('&&s[j] == ')') return 1;if (s[i] == '['&&s[j] == ']') return 1;return 0;}int main(){while (~scanf("%s", s + 1)){if (s[1] == 'e') break;ms(dp, 0);int n = strlen(s + 1);for (int i = n; i > 0; i--){for (int j = i + 1; j <= n; j++){if (check(i, j)) dp[i][j] = dp[i + 1][j - 1] + 2;for (int k = i; k <= j; k++)dp[i][j] = max(dp[i][j], dp[i][k] + dp[k][j]);}}printf("%d\n", dp[1][n]);}}