【POJ2955】【Brackets】

Brackets

Time Limit: 1000MS Memory Limit: 65536KTotal Submissions: 4257 Accepted: 2259

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

66406

Source

Stanford Local 2004

分为2种情况

1. 可以拆分 dp[i][j] = dp[i][k] + dp[k+1][j] 

2.可以不拆分  dp[i][j] = dp[i+1][j-1]

开始的时候忘记了可以不拆分的情况，，，。 因为当时思路一直在想如何合并啊，然后还企图尝试使用3个数组l[][] r[][] dp[][]，然后无法实现告终。。。后来才知道这道题想着如何拆分比较好。。。

#include <iostream>#include <cstring>#include <cmath>#include <queue>#include <stack>#include <list>#include <map>#include <set>#include <string>#include <cstdlib>#include <cstdio>#include <algorithm>using namespace std;    char s[110];int dp[110][110];int len;bool match(int i,int j){if(s[i] == '[' && s[j] == ']' || s[i] == '(' && s[j] == ')')return true;return false;}int main(){    while(gets(s)){if(strcmp(s,"end") == 0) break;    memset(dp,0,sizeof(dp));len = strlen(s);for(int i=0;i+1<len;i++){if(match(i,i+1)) dp[i][i+1] = 1;else dp[i][i+1] = 0;}for(int l=3;l<=len;l++) {for(int i=0;i+l-1<len;i++){int j = i + l -1;if(match(i,j)) dp[i][j] = max(dp[i][j],dp[i+1][j-1] + 1);for(int k=1;k<l;k++){dp[i][j] = max(dp[i][j],dp[i][i+k-1] + dp[i+k][j]);}}}printf("%d\n",dp[0][len-1]*2);}    return 0;}