POJ__2955 Brackets

题目描述：

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 [([])].

给你一串括号，问你括号匹配的数目,输入end时结束。

样例：
input:

((()))
()()()
([]])
)[)(
([][][)
end

output:

6
6
4
0
6

模板式的区间DP，选取区间长度，选取起点然后确定终点，暴力循环更新，状态转移方程：dp[i][j]=dp[i-1][j+1]+2，DP数组储存区间[i,j]内匹配括号的数目

AC代码：

#include<iostream>
#include<cstdio>
#include<cstring>
#include<string>
#include<cmath>
#include<stack>
#include<queue>
#include<vector>
#include<algorithm>
using namespace std;
const int M=105;
const int INF=1e9+1;
char s[M];
int dp[M][M];


bool check(char a,char b)
{
    if (a=='('&&b==')') return true;
    if (a=='['&&b==']') return true;
    return false;
}
int main()
{
    while(gets(s)!=NULL)
    {
        if(!strcmp(s, "end")) break;
        int len=strlen(s);
        for (int i=0;i<len;i++)
        {
            if (check(s[i],s[i+1]))
                dp[i][i+1]=2;
            else
                dp[i][i+1]=0;
        }
        for (int l=2;l<len;l++)
        {
            for (int i=0;i+l<len;i++)
            {
                int r=i+l;
                dp[i][r]=0;
                if (check(s[i],s[r]))
                    dp[i][r]=dp[i+1][r-1]+2;
                for (int k=i;k<r;k++)
                dp[i][r]=max(dp[i][r],dp[i][k]+dp[k+1][r]);

            }
        }
        printf("%d\n",dp[0][len-1]);
    }
    return 0;
}