POJ 2955

Brackets

Time Limit: 1000MS Memory Limit: 65536KTotal Submissions: 9255 Accepted: 4947

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

66406

Source

Stanford Local 2004

题意：

输出给你的括号匹配最长长度，可以不连续。

POINT：

看代码里的状态转移方程

#include <iostream>#include <stdio.h>#include <string.h>#include <vector>#include <queue>using namespace std;const int maxn = 111;int dp[maxn][maxn];int main(){    char s[maxn];    while(~scanf("%s",s+1)){        if(s[1]=='e') break;        int n=(int)strlen(s)-1;        memset(dp,0,sizeof dp);        for(int len=2;len<=n;len++){            for(int i=1;i<=n-len+1;i++){                int j=i+len-1;                if((s[i]=='('&&s[j]==')')||(s[i]=='['&&s[j]==']')) dp[i][j]=dp[i+1][j-1]+2;                for(int k=1;k<j;k++)                    dp[i][j]=max(dp[i][j],dp[i][k]+dp[k+1][j]);            }        }        printf("%d\n",dp[1][n]);    }}