POJ2955-Brackets

最新推荐文章于 2019-03-14 09:30:44 发布

Wang_128

最新推荐文章于 2019-03-14 09:30:44 发布

阅读量317

点赞数

CC 4.0 BY-SA版权

分类专栏： POJ ----区间dp

本文链接：https://blog.youkuaiyun.com/a664607530/article/details/52557985

POJ 同时被 2 个专栏收录

159 篇文章

订阅专栏

----区间dp

11 篇文章

订阅专栏

本文介绍了一种解决最长有效括号子序列问题的方法，通过区间动态规划算法找到给定字符串中最长的有效括号序列长度。

摘要生成于 C知道，由 DeepSeek-R1 满血版支持，前往体验 >

Brackets

Time Limit: 1000MS		Memory Limit: 65536K
Total Submissions: 6517		Accepted: 3492

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

Source

Stanford Local 2004

题意：求出互相匹配的括号最多的数量

解题思路：一道区间DP，dp[i][j]存的是i~j区间内匹配的个数

#include <iostream>
#include <cstdio>
#include <cstring>
#include <stack>
#include <queue>
#include <cmath>
#include <algorithm>

using namespace std;

int check(char a,char b)
{
    if(a=='('&&b==')') return 1;
    else if(a=='['&&b==']') return 1;
    else return 0;
}

int main()
{
    char ch[120];
    int dp[120][120];
    while(~scanf("%s",ch)&&strcmp(ch,"end"))
    {
        memset(dp,0,sizeof dp);
        int len=strlen(ch);
        for(int i=1;i<=len;i++)
        {
            for(int j=i-1;j>0;j--)
            {
                if(check(ch[j-1],ch[i-1]))
                    dp[j][i]=dp[j+1][i-1]+2;
                else
                    dp[j][i]=max(dp[j+1][i],dp[j][i-1]);
                for(int k=j;k<=i;k++)
                    dp[j][i]=max(dp[j][i],dp[j][k]+dp[k][i]);
            }
        }
        printf("%d\n",dp[1][len]);
    }
    return 0;
}