Poj 2955 Brackets【区间dp】

Brackets

Time Limit: 1000MS Memory Limit: 65536KTotal 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

6

6

4

0

6

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]);    }}