http://poj.org/problem?id=2955

?

?

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?…?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

6
6
4
0
6

?

?p[i][j]表示從i到j個可以組成的括號最大值，則若dp[i+1][j]已取到最大值，則dp[i][j] 的取值為 dp[i+1][j] ， 或若 s[i] 與 第i+1個到第j個中某個括號匹配（假定為第k個），則有dp[i][j] = max(dp[i+1][j], dp[i+1][k-1] + 2 + dp[k+1][j]) （注：要考慮k == i+1的情況要分開討論)

?

#include<algorithm>
#include<iostream>
#include<cstring>
#include<cstdio>
#include<cmath>
using namespace std;const int INF = 0x3f3f3f3f;
#define N 105char s[N];
int  dp[N][N];int main()
{while(scanf("%s", s), strcmp(s, "end")){int i, j, k, len=strlen(s)-1;memset(dp, 0, sizeof(dp));for(i=len-1; i>=0; i--){for(j=i+1; j<=len; j++){dp[i][j] = dp[i+1][j];for(k=i+1; k<=j; k++){if((s[i]=='(' && s[k]==')') || (s[i]=='[' && s[k]==']')){if(k==i+1) dp[i][j] = max(dp[i][j], dp[k+1][j]+2);else       dp[i][j] = max(dp[i][j], dp[i+1][k-1]+dp[k+1][j]+2);}}}}printf("%d\n", dp[0][len]);}return 0;
}

?

記憶化索搜：

（感覺記憶化搜索只是把在遞歸中已經計算過的值給記錄下來， 不知道是否理解有悟，慢慢用吧！！！）

#include<stdio.h>
#include<string.h>
#include<stdlib.h>#define N 105
#define max(a,b) (a>b?a:b)char s[N];
int  dp[N][N];int OK(int L, int R)
{if((s[L]=='[' && s[R]==']') || (s[L]=='(' && s[R]==')'))return 2;return 0;
}int DFS(int L, int R)
{int i;if(dp[L][R]!=-1)return dp[L][R];if(L+1==R)return OK(L,R);if(L>=R)return 0;dp[L][R] = DFS(L+1, R);for(i=L+1; i<=R; i++){if(OK(L,i))dp[L][R] = max(dp[L][R], DFS(L+1, i-1)+DFS(i+1, R)+2);}return dp[L][R];
}int main()
{while(scanf("%s", s), strcmp(s, "end")){memset(dp, -1, sizeof(dp));printf("%d\n", DFS(0, strlen(s)-1));}return 0;
}

?

轉載于:https://www.cnblogs.com/YY56/p/5051617.html

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