POJ2955-Brackets（区间dp 括号匹配）

题目链接

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

---

题意

给了一个括号匹配的定义，如果 ( ) [ ( ) ] 都是匹配的，但是 [ ( ] ( ) 就是不匹配的。给一个字符串，问满足括号匹配的子串的最大长度是多少。

思路

设dp[ i ] [ j ] 是字符串i 到 j 的括号匹配的最大长度，从子情况得到父情况主要有两种。

① ( ... ) [ ... ] 型，即左右分开，合起来是dp[ i ] [ j ]，用k循环 求所有的最大值 ；

② ( ....  ) 型，即最左和最后是匹配的，中间的情况加一是dp[ i ] [ j ] .

所有情况比较，求最大值，就是dp[ i ] [ j ].

---

#include <cstdio>
#include <iostream>
#include <algorithm>
#include <cstring>
#define ll long long
#define inf 0x3f3f3f3f
using namespace std;

int dp[110][110];
char str[110];

int main()
{
    while(~scanf("%s",str+1))
    {
        if(str[1]=='e') break;
        memset(dp,0,sizeof(dp));

        int len = strlen(str+1);
        for(int l=1;l<=len;l++){
            for(int i=1;i<=len;i++){
                int j = i+l-1;
                if(j>len) continue;
                if((str[i]=='('&&str[j]==')')||(str[i]=='['&&str[j]==']') )
                    dp[i][j] = dp[i+1][j-1] + 2;
                    //dp[i+1][j-1],当i+1>j-1,值仍为0
                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][len]);
    }
    return 0;
}