poj--2955--Brackets（区间dp）

Brackets

Time Limit: 1000MS Memory Limit: 65536KB 64bit IO Format: %I64d & %I64u

SubmitStatus

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 ofs. That is, you wish to find the largest m such that for indicesi₁, 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

区间dp，每次在i--j寻找最大的匹配数，更新dp[i][j]，每一步都应该有dp[i][j]=dp[i+1][j]，因为下一步循环不一定能进入

#include<stdio.h>#include<string.h>#include<algorithm>using namespace std;char str[10010];int dp[1010][1010];bool judge(char a,char b){if(a=='('&&b==')') return true;if(a=='['&&b==']') return true;return false;}int main(){while(scanf("%s",str)!=EOF){if(strcmp(str,"end")==0) break;int len=strlen(str);memset(dp,0,sizeof(dp));for(int i=len-1;i>=0;i--){for(int j=i;j<len;j++){dp[i][j]=dp[i+1][j];for(int k=i+1;k<=j;k++)if(judge(str[i],str[k]))dp[i][j]=max(dp[i][j],dp[i+1][k-1]+1+dp[k+1][j]);}}memset(str,'\0',sizeof(str));printf("%d\n",dp[0][len-1]*2);}return 0;}