poj 2955

Brackets

Time Limit: 1000MS		Memory Limit: 65536K
Total Submissions: 3961		Accepted: 2085

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 …
an, 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, …, im where 1 ≤
i1 < i2 < … < im ≤
n, ai1ai2 …
aim 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

题目链接：http://poj.org/problem?id=2955

【题意】最多多少个匹配的

【思路】区间dp，dp[i][j]表示第i个到第j个最多多少个匹配.

                                   dp[i][j]=min(dp[i][j],dp[i+1][j-1]+1)  s[i]==s[j];

                                   dp[i][j]=min(dp[i][j],dp[i]][k]+dp[k+1][j]) i<k<j;

【代码】

/*************************************************************************
> File Name: poj2955.cpp
> Author: wanghao
> Mail: haohaoac@163.com
> Created Time: 2015年07月09日 星期四 22时52分19秒
************************************************************************/

#include<iostream>
#include<cstring>
#include<cstdio>
#define ll long long
using namespace std;

int dp[110][110];
int judge(char a,char b)
{
if(a=='('&&b==')')
return 1;
if(a=='['&&b==']')
return 1;
return 0;
}
int main()
{
char s[110];
while(scanf("%s",s+1)!=EOF)
{
if(strcmp(s+1,"end")==0)
break;
int len=strlen(s+1);
memset(dp,0,sizeof(dp));
for(int l=1;l<=len;l++)
{
for(int i=1;i+l-1<=len;i++)
{
int j=i+l-1;
if(judge(s[i],s[j]))
dp[i][j]=dp[i+1][j-1]+1;
for(int k=i;k<=j;k++)
dp[i][j]=max(dp[i][k]+dp[k+1][j],dp[i][j]);

}
}
cout<<2*dp[1][len]<<'\n';
}
return 0;
}