（POJ 2955）Brackets 区间DP 最大括号匹配

Brackets
Time Limit: 1000MS Memory Limit: 65536K
Total Submissions: 6647 Accepted: 3577
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
Source

Stanford Local 2004

题意：
给你由(,),[,]组成的字符串，从中取出一子串，问最大的括号匹配数？

分析：
求解DP的问题要从三个方面出发：子问题+最优解+极限状态。

设dp[i][j]表示字符串str[i]~str[j]的最大匹配数。

子问题：
（1）外层括号可以包含内层括号，所以当str[i],str[j]匹配时，dp[i][j] = dp[i+1][j-1] + 2;
（2）dp[i][j] 可以被分为很多不同的段的解的和值，即dp[i][j] = dp[i][k]+dp[k+1][j];

最优解：
dp[i][j] 为上面所有情况的最大值。

极限状态：
长度为1时：dp[i][i] = 0; 然后从长度为2~n开始递推出所有解。

AC代码：

#include <iostream>#include <cstdio>#include <cstring>#include <string>#include <algorithm>using namespace std;char str[110];int dp[110][110];int main(){    while(scanf("%s",str) && strcmp(str,"end"))    {        int n = strlen(str);        memset(dp,0,sizeof(dp));        for(int len=2;len<=n;len++)        {            for(int i=0;i+len<=n;i++)            {                int j = i+len-1;                if(str[i] == '(' && str[j] == ')' || str[i] == '[' && str[j] == ']')                    dp[i][j] = max(dp[i][j],dp[i+1][j-1]+2);                for(int k=i;k<j;k++)                    dp[i][j] = max(dp[i][j],dp[i][k]+dp[k+1][j]);            }        }        cout<<dp[0][n-1]<<endl;    }    return 0;}