ACM POJ 2955 题解 区间DP

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Brackets。区间划分DP，没有限制的区间划分。详见本文内容

题解

Brackets

限制

Time Limit: 1000MS
Memory Limit: 65536K

描述

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 $a_1a_2…a_n$, 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 $i_1, i_2, …, i_m$ where $1 \le i_1 \lt i_2 \lt … \lt i_m \le n$, $a_{i_1}a_{i_2} … a_{i_m}$ is a regular brackets sequence.
Given the initial sequence ([([]])], the longest regular brackets subsequence is [([])].

输入格式

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.

输出格式

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

样本

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((()))
()()()
([]])
)[)(
([][][)
end
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0
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思路

区间划分DP，没有限制的区间划分。
dp[a][b]存的是区间[a,b]内满足要求的符号表达式的最大长度。
然后dp[a][b]的最大长度，是对区间[a,b]进行划分得到的所有结果中，的最大值。
如果str[a]==’(‘并且str[b]==’)’或者是str[a]==’[‘并且str[b]==’]’，即a位置与b位置正好能够配对的时候，dp[a][b]还要与dp[a+1][b-1]作比较，dp[a][b]取最大值。
解释一下代码，
16行的i是对区间长度进行迭代。17行的j是对不同的区间起始节点进行迭代。18行是对[j,j+i]区间的不同划分点进行迭代。

代码

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#include <iostream>
#include <cstdio>
#include <cstring>

using namespace std;

int dp[105][105];
char str[105];

int main(){
    while(~scanf("%s", str)){
        if(str[0] == 'e') break;
        int n = strlen(str);
        memset(dp, 0, sizeof dp);

        for(int i = 0; i < n; i++){
            for(int j = 0; j < n-i; j++){
                for(int k = j; k < j+i; k++){
                    dp[j][j+i] = max(dp[j][j+i], dp[j][k]+dp[k+1][j+i]);
                }
                if((str[j] == '(' && str[j+i] == ')') || (str[j] == '[' && str[j+i] == ']'))
                    dp[j][j+i] = max(dp[j][j+i], dp[j+1][j+i-1]+2);
            }
        }

        printf("%d\n", dp[0][n-1]);
    }
    return 0;
}