POJ：2955-Brackets（经典：括号匹配）

传送门：http://poj.org/problem?id=2955

Brackets

Time Limit: 1000MS Memory Limit: 65536K

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

---

1. 这是一个区间dp的经典例题，刚开始的时候还以为可以模拟做，但是事实就是这是不行的，只是自己想的太简单，括号的匹配有很多种。
2. 区间dp执行的时候，里面先是一段区间一段区间的跑，然后枚举这个区间里面每一种匹配情况，由于在选择区间的时候是从小开始选择，所以在枚举每一种匹配情况的时候都是从前面得到的答案来转移的，这也体现了动态规划，从子问题转移。只不过区间dp里面是从一个小的区间慢慢转移到整个区间。但是怎么处理区间里面的状态要看当前问题的特点。

---

#include<stdio.h>
#include<string.h>

using namespace std;
const int maxn = 110;
char s[maxn];
int dp[maxn][maxn];

bool checke(int S,int E)
{
    if(s[S] == '(' && s[E] == ')')
        return true;
    if(s[S] == '[' && s[E] == ']')
        return true;
    return false;
}

int main()
{
    while(scanf("%s",s))
    {
        if(strcmp(s,"end") == 0)
            break;
        memset(dp,0,sizeof(dp));
        int len = strlen(s);
        for(int i=1;i<len;i++)
            for(int j=0,k=i;k<len;k++,j++)
            {
                if(checke(j,k))
                    dp[j][k] = dp[j+1][k-1] + 2;
                for(int z=j;z<k;z++)
                    if(dp[j][z] + dp[z+1][k] > dp[j][k])
                        dp[j][k] = dp[j][z] + dp[z+1][k];
            }
        printf("%d
",dp[0][len-1]);
    }
    return 0;
}