poj 2955 Brackets（区间dp）

最新推荐文章于 2019-07-10 00:08:39 发布

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最新推荐文章于 2019-07-10 00:08:39 发布

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分类专栏：动态规划文章标签： acm dp

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本博客探讨了一个仅包含括号字符的字符串中，寻找最长合法括号匹配子序列的问题。通过动态规划方法，实现高效求解并提供代码实现。

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题目链接

Brackets

Time Limit: 1000MS		Memory Limit: 65536K
Total Submissions: 3470		Accepted: 1792

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

题意：给一个只包含括号的串，问该串的最长的合法括号匹配的子序列的长度？

题解：先考虑第一个括号，有两种操作：删除这个括号；将该括号与另一个括号匹配。

用dp[l][r]表示区间[l,r]的最长合法子序列的长度，转移就是：

1，删除第一个括号

dp[l][r]=dp[l+1][r]

2，将第一个括号和第i个括号匹配

dp[l][r]=min(dp[l+1][i-1]+dp[i+1][r]+2);

代码如下：

#include<stdio.h>
#include<iostream>
#include<string.h>
#include<algorithm>
#include<string>
#include<math.h>
#define nn 110
#define inff 0x7fffffff
#define eps 1e-8
typedef long long LL;
const LL inf64=LL(inff)*inff;
using namespace std;
int n;
char s[nn];
int dp[nn][nn];
int lj[nn][nn];
bool check(int l,int r)
{
    if(s[l]=='('&&s[r]==')')
        return true;
    if(s[l]=='['&&s[r]==']')
        return true;
    return false;
}
int dfs(int l,int r)
{
    if(dp[l][r]!=-1)
        return dp[l][r];
    if(l>r)
        return 0;
    if(l==r)
        return dp[l][r]=0;
    dp[l][r]=0;
    int i;
    for(i=l+1;i<=r;i++)
    {
        if(check(l,i))
            dp[l][r]=max(dp[l][r],dfs(l+1,i-1)+dfs(i+1,r)+2);
    }
    dp[l][r]=max(dp[l][r],dfs(l+1,r));
    return dp[l][r];
}
int main()
{
    int i;
    while(scanf("%s",s)!=EOF)
    {
        if(strcmp(s,"end")==0)
            break;
        int len=strlen(s);
        memset(dp,-1,sizeof(dp));
        printf("%d\n",dfs(0,len-1));
    }
    return 0;
}