POJ 2955 区间dp

Brackets
Time Limit: 1000MS   Memory Limit: 65536K
Total Submissions: 2446   Accepted: 1258

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 a1a2an, 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, ai1ai2aim 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


题意:求最长合法括号序列,

两种写法:

/* ***********************************************
Author :rabbit
Created Time :2014/3/10 19:03:58
File Name :A.cpp
************************************************ */
#pragma comment(linker, "/STACK:102400000,102400000")
#include 
#include 
#include 
#include 
#include 
#include 
#include 
#include 
#include 
#include 
#include 
#include 
#include 
#include 
using namespace std;
#define INF 0x3f3f3f3f
#define eps 1e-8
#define pi acos(-1.0)
typedef long long ll;
int dp[300][300];
char str[400];
bool match(char a,char b){
	if(a=='('&&b==')')return 1;
	if(a=='['&&b==']')return 1;
	return 0;
}
int main(){
	while(~scanf("%s",str)&&str[0]!='e'){
		int len=strlen(str);
		memset(dp,0,sizeof(dp));
		for(int i=0;i+1

/* ***********************************************
Author :rabbit
Created Time :2014/3/10 19:03:58
File Name :A.cpp
************************************************ */
#pragma comment(linker, "/STACK:102400000,102400000")
#include 
#include 
#include 
#include 
#include 
#include 
#include 
#include 
#include 
#include 
#include 
#include 
#include 
#include 
using namespace std;
#define INF 0x3f3f3f3f
#define eps 1e-8
#define pi acos(-1.0)
typedef long long ll;
int dp[300][300];
char str[400];
int dfs(int l,int r){
	int &res=dp[l][r];
	if(res!=-1)return res;
	if(r-l<2)return res=0;
	for(int i=l+1;i


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