POJ 2955 Brackets (区间DP)

Brackets
Time Limit: 1000MS Memory Limit: 65536K
Total Submissions: 6420 Accepted: 3436
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

给出一个串，请取出其中的一些字符组成的新串括号必须是两两匹配的。可以不连续地取，但必须保持先后顺序。

#include "cstring"#include "cstdio"#include "iostream"#include "string.h"#include "algorithm"using namespace std;char s[1005];int dp[1005][1005];bool check(int left,int right){    if(s[left]=='('&&s[right]==')')        return true;    if(s[left]=='['&&s[right]==']')        return true;    return false;}int main(){    while(scanf("%s",s)&&strcmp(s,"end")!=0)    {        int len=strlen(s);        memset(dp,0,sizeof(dp));        //l表示区间长度。区间的长度可以从1~len        for(int l=1;l<=len;l++)        {            //枚举该长度下的所有区间            for(int i=0;i+l-1<len;i++)            {                //若区间守卫可以构成一个匹配，那么答案就+2                if(check(i,i+l-1))                    dp[i][i+l-1]=dp[i+1][i+l-2]+2;                //合并区间                for(int j=i;j<i+l-1;j++)                    dp[i][i+l-1]=max(dp[i][i+l-1],dp[i][j]+dp[j+1][i+l-1]);            }        }        printf("%d\n",dp[0][len-1]);    }}