POJ2955 Brackets【区间DP 基础】

本文主要是介绍POJ2955 Brackets【区间DP 基础】，希望对大家解决编程问题提供一定的参考价值，需要的开发者们随着小编来一起学习吧！

Brackets

Time Limit: 1000MS		Memory Limit: 65536K
Total Submissions: 12763		Accepted: 6776

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

问题链接：POJ2955 Brackets

问题描述：求最大括号匹配

解题思路：基础的区间DP，dp[i][j]表示 [i,j] 区间内的最大匹配数，如果s[i]和s[j]匹配了，那么dp[i][j]=dp[i+1][j-1]+2。接着在区间[i,j]内设置断点k (i<=k<j) dp[i][j]=max(dp[i][j],dp[i][k]+dp[k+1][j])

AC的C++程序：

#include<iostream>
#include<cstring>using namespace std;using namespace std;const int N=105;
int dp[N][N];int main()
{char s[N];while(~scanf("%s",s)){if(s[0]=='e')break;memset(dp,0,sizeof(dp));int n=strlen(s);for(int len=1;len<n;len++)for(int i=0,j=len;j<n;i++,j++){if((s[i]=='('&&s[j]==')')||(s[i]=='['&&s[j]==']'))dp[i][j]=dp[i+1][j-1]+2;//加2是由于s[i]和s[j]两个匹配了for(int k=i;k<j;k++)//遍历各个断点 dp[i][j]=max(dp[i][j],dp[i][k]+dp[k+1][j]); }printf("%d\n",dp[0][n-1]);}return 0;
}

 

这篇关于POJ2955 Brackets【区间DP 基础】的文章就介绍到这儿，希望我们推荐的文章对编程师们有所帮助！

---