POJ2955-Brackets

Brackets

Time Limit: 1000MS Memory Limit: 65536KTotal Submissions: 6517 Accepted: 3492

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：求出互相匹配的括号最多的数量

解题思路：一道区间DP，dp[i][j]存的是i~j区间内匹配的个数

#include <iostream>#include <cstdio>#include <cstring>#include <stack>#include <queue>#include <cmath>#include <algorithm>using namespace std;int check(char a,char b){    if(a=='('&&b==')') return 1;    else if(a=='['&&b==']') return 1;    else return 0;}int main(){    char ch[120];    int dp[120][120];    while(~scanf("%s",ch)&&strcmp(ch,"end"))    {        memset(dp,0,sizeof dp);        int len=strlen(ch);        for(int i=1;i<=len;i++)        {            for(int j=i-1;j>0;j--)            {                if(check(ch[j-1],ch[i-1]))                    dp[j][i]=dp[j+1][i-1]+2;                else                    dp[j][i]=max(dp[j+1][i],dp[j][i-1]);                for(int k=j;k<=i;k++)                    dp[j][i]=max(dp[j][i],dp[j][k]+dp[k][i]);            }        }        printf("%d\n",dp[1][len]);    }    return 0;}