Poj 2955 Brackets【区间dp】

最新推荐文章于 2020-05-17 17:24:15 发布

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最新推荐文章于 2020-05-17 17:24:15 发布

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分类专栏： dp 文章标签： Poj 2955 pku 2955

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本文介绍了一种使用动态规划算法解决寻找字符串中最长有效括号匹配的问题，通过实例详细解析了算法实现过程。

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Brackets

Time Limit: 1000MS		Memory Limit: 65536K
Total Submissions: 6105		Accepted: 3269

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

题目大意:求给出的字符串的最长匹配。

思路：

1、设定dp【i】【j】表示从i到j这个区间内字符串的最长匹配。

2、那么不难理解：dp【i】【j】=dp【i+1】【j-1】+2（if（j>i&&i位子上的字符和j位子上的字符能够匹配上。））

3、如果有这样的字符串：（）（），其中明显有dp【1】【2】=2，dp【3】【4】=2；dp【1】【4】=2，因为两个匹配是可以合并的,如：dp【1】【4】=dp【1】【2】+dp【3】【4】=4所以还有：

dp【i】【j】=max（dp【i】【j】，dp【i】【k】+dp【k+1】【j】）；

Ac代码：

#include<stdio.h>
#include<string.h>
#include<iostream>
using namespace std;
int dp[150][150];
int main()
{
    char a[150];
    while(~scanf("%s",a))
    {
        memset(dp,0,sizeof(dp));
        if(strcmp(a,"end")==0)
        {
            break;
        }
        int n=strlen(a);
        for(int d=1;d<n;d++)
        {
            for(int i=0;i+d<n;i++)
            {
                int j=i+d;
                if(a[j]==')'&&a[i]=='('||a[j]==']'&&a[i]=='[')
                {
                    dp[i][j]=dp[i+1][j-1]+2;
                }
                for(int k=0;k<n;k++)
                {
                    dp[i][j]=max(dp[i][j],dp[i][k]+dp[k+1][j]);
                }
            }
        }
        printf("%d\n",dp[0][n-1]);
    }
}