POJ2955Brackets(区间DP模板题)

Brackets

Time Limit: 1000MS Memory Limit: 65536KTotal Submissions: 7556 Accepted: 4020

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

题目大意:给你一串字符串，然后求出来这个字符串中，最长的符号匹配串，也就是说(()),[[]],[][]这样是匹配的，而这样[[],(][))就不是

解题思路:这是个区间DP，dp[j][k]表示j到k匹配长度，要是dp[j]==dp[k]相同，意味着dp[j][k]=dp[j+1][k-1]+2

#include<iostream>    #include<cstdio>  #include<stdio.h>  #include<cstring>    #include<cstdio>    #include<climits>    #include<cmath>   #include<vector>  #include <bitset>  #include<algorithm>    #include <queue>  #include<map>  using namespace std;int main(){int i, j, u, k, n;int dp[105][105];for (;;){string str;cin >> str;if (str == "end"){return 0;}memset(dp, 0, sizeof(dp));n = str.length();for (i = 1; i < n; i++)//区间长度{for (j = 0,k=i; k < n; k++,j++)//区间头尾位置{if (str[j] == '['&&str[k] == ']' || str[j] == '('&&str[k] == ')'){dp[j][k] = dp[j + 1][k - 1] + 2;}for (u = j; u <=k; u++){dp[j][k] = max(dp[j][u] + dp[u+1][k], dp[j][k]);}}}cout << dp[0][n - 1] << endl;}}