poj2955 Brackets--最大括号匹配数【区间DP】

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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：

      找出括号匹配的最长长度。

分析：

       （之前做过的一道区间DP都忘了，详细写写，理解有限，大佬勿喷，欢迎大佬指正）；

       理解：区间DP主要是将大区间拆成几个区间，先求小区间的最优值，然后合并区间求出区间最优解。

一般是先根据区间的的长度，枚举区间的起点，然后构造状态转移方程。

       本题：设dp[i][j]为区间i到j的最大括号匹配，枚举区间的长度，当出现配对的括号时，dp[i][j]=dp[i-1][j+1]+2，枚举区间的分割点，构造求最大值的状态转移方程，dp[i][j]=max(dp[i][j],dp[i][k]+dp[k][j])。-

代码：

#include<cstdio>
#include<cstring>
#include<algorithm>
#include<iostream>
#include<cmath>
using namespace std;
char s[108];
int dp[108][108];
int main()
{while(scanf("%s",s+1)==1&&s[1]!='e'){int n=strlen(s+1);memset(dp,0,sizeof(dp));for(int len=2;len<=n;len++)///区间长度{for(int i=1;i<=n;i++)///枚举起点{int j=i+len-1;if(j>n) break;if((s[i]=='('&&s[j]==')')||((s[i]=='['&&s[j]==']'))){dp[i][j]=dp[i+1][j-1]+2;}for(int k=i;k<j;k++)///枚举分割点，构造状态转移方程{dp[i][j]=max(dp[i][j],(dp[i][k]+dp[k][j]));}}}printf("%d\n",dp[1][n]);}return 0;
}

 

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