POJ 2955-Brackets（括号匹配-区间DP）
Brackets
| Time Limit: 1000MS | Memory Limit: 65536K | |
| Total Submissions: 5484 | Accepted: 2946 |
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 … an, 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, ai1ai2 … aim 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
Source
题目意思：
最大括号匹配数。解题思路：
From定义dp [ i ] [ j ] 为串中第 i 个到第 j 个括号的最大匹配数目
那么我们假如知道了 i 到 j 区间的最大匹配，那么i+1到 j+1区间的是不是就可以很简单的得到。
那么 假如第 i 个和第 j 个是一对匹配的括号那么 dp [ i ] [ j ] = dp [ i+1 ] [ j-1 ] + 2 ;
那么我们只需要从小到大枚举所有 i 和 j 中间的括号数目，然后满足匹配就用上面式子dp，然后每次更新dp [ i ] [ j ]为最大值即可。
更新最大值的方法是枚举 i 和 j 的中间值，然后让 dp[ i ] [ j ] = max ( dp [ i ] [ j ] , dp [ i ] [ f ] + dp [ f+1 ] [ j ] ) 。
#include
#include
#include
#include
#include
#include
#include
using namespace std;
#define MAXN 202
#define INF 999999
int dp[MAXN][MAXN];
int main()
{
string s;
while(cin>>s)
{
if(s=="end") break;
memset(dp,0,sizeof(dp));
int i,j,k,f,len=s.size();
for(i=1; i