POJ 2955- Brackets[区间dp]

最新推荐文章于 2019-08-03 11:55:32 发布

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最新推荐文章于 2019-08-03 11:55:32 发布

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分类专栏： ACM 文章标签： ACM poj 区间dp

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本文介绍了一种使用动态规划算法来解决寻找给定字符串中最长的有效括号子序列的问题，并给出了具体的实现代码。

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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 … 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

思路: 对左括号不进行处理, rev数组存储一个括号的反括号如rev['('] = ')' 其余的同理

dp[i][j] 代表的是i到j位置匹配的最大个数

转移方程dp[i][j] = dp[i][j-1], 对任意i,j

dp[i][j] = (dp[i][k] + dp[k+1][j] +1, a[k] = rev[a[j]]

answer = dp[0][N-1] *2

#include<iostream>
#include<algorithm>
#include<cmath>
#include<vector>
#include<queue>
#include<iomanip>
#include<stdlib.h>
#include<cstdio>
#include<string>
#include<string.h>
#include<set>
#include<stack>
#include<map>
using namespace std;

#define rep(i, n) for(int i=0; i<n ;i++)
#define rep1(i, n) for(int i=1; i<=n; i++)
typedef long long ll;
typedef  pair<int,int> P;
const int INF = 0x7fffffff;
const int MAX_N = 105;
const int MAX_V = 0;
const int MAX_M = 0;
const int MAX_Q = 1e5+5;	

int  dp[MAX_N][MAX_N];
int l[3] = {'(','[','{'}, r[3] = {')',']','}'};
int rev[256];

int main(){
	string a;
	for(int i=0; i<3; i++){
		rev[l[i]] = r[i];
		rev[r[i]] = l[i];
	}
	while(getline(cin, a)){
		if(a=="end") break;
		int N = a.length();
		for(int i=0; i<N; i++){
			dp[i][i] = 0;
		}
		for(int j=1; j<N; j++){
			for(int i=0; i<j; i++){
				dp[i][j] = dp[i][j-1];
				bool flag = false;
				for(int s =0; s<3; s++)
					if(a[j]==r[s])
						flag = true;
				if(flag){
					for(int k=i; k<j; k++){
						if(rev[a[k]]==a[j]){
							dp[i][j] = max(dp[i][j], dp[i][k]+dp[k+1][j-1]+1);
						}
					}
				}
			}
		}	
		cout<<dp[0][N-1]*2<<endl;
	}
}