POJ-2955括号匹配问题（区间DP）

Brackets

Time Limit: 1000MS		Memory Limit: 65536K
Total Submissions: 4834		Accepted: 2574

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

6　　　　　　　　　 

6 4 0 6

Difficulty: （1）.状态：dp[i][j]为i~j的最大括号数。
   
            （2）. 转移：考虑第i个括号，有两种情况：

                      1.i无效，直接算dp[i + 1][j];

                      2.找到和i匹配的右括号k，分两边算并加起来。dp[i][j] = dp[i+1][k-1] + 2 + dp[k + 1][j]
感想：记忆化搜索实质上就是暴力枚举。

#include <cstdio>
#include <iostream>
#include <cstdlib>
#include <algorithm>
#include <ctime>
#include <cmath>
#include <string>
#include <cstring>
#include <stack>
#include <queue>
#include <list>
#include <vector>
#include <map>
#include <set>
using namespace std;

const int INF=0x3f3f3f3f;
const double eps=1e-10;
const double PI=acos(-1.0);
#define maxn 1100

char a[maxn];
int dp[maxn][maxn];
int is(char b, char c)
{
    if(b == '(' && c == ')' || b == '[' && c == ']')
        return 1;
    else
        return 0;

}
int dfs(int st, int ed)
{
    //
    if(st > ed)
        return 0;
    if(st == ed)
       return 0;
    if(dp[st][ed] != -1) return dp[st][ed];
    int res = dfs(st+1, ed);
    for(int k = st+1; k <= ed; k++)
        if(is(a[st],a[k]))
        {
            res = max(res,dfs(st+1,k-1) + 2 + dfs(k+1,ed));
            flag = 1;
        }
        dp[st][ed] = res;
    return dp[st][ed];
}
int main()
{
    while(~scanf("%s", a))
    {
        if(strcmp(a, "end") == 0)
            break;
        memset(dp, -1, sizeof dp);
        int ed = strlen(a)-1;
        printf("%d
", dfs(0, ed));
    }
    return 0;
}