poj 2955 Brackets（区间dp）

                                                                                                Brackets

Time Limit: 1000MS		Memory Limit: 65536K
Total Submissions: 7756		Accepted: 4114

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

Source

Stanford Local 2004

以前写过一道类似的题 最优矩阵乘法的那个 也是一道dp 发现这叫区间dp

dp[i][j]表示【i j]这个区间的最大匹配个数

我们可以通过求【i，k],[k+1,j]来求这个【i,j]

所以我们只要枚举这个区间找到最大的那个

转移方程 dp[i][j]=max(dp[i][k],dp[k+1][j]);

#include<iostream>
#include<cstdio>
#include<cstdlib>
#include<cctype>
#include<cmath>
#include<cstring>
#include<map>
#include<stack>
#include<set>
#include<vector>
#include<algorithm>
#include<string.h>
typedef long long ll;
typedef unsigned long long LL;
using namespace std;
const int INF=0x3f3f3f3f;
const double eps=0.0000000001;
const int N=1000+10;
char str[N];
int dp[N][N];
int judge(int x,int y){
    if(str[x]=='('&&str[y]==')')return 1;
    if(str[x]=='['&&str[y]==']')return 1;
    return 0;
}
int main(){
    while(gets(str)){
        if(strcmp(str,"end")==0)break;
        memset(dp,0,sizeof(dp));
        int len=strlen(str);
        for(int t=1;t<len;t++){
            for(int i=0;i+t<len;i++){
                int j=i+t;
                if(judge(i,j)){
                    if(j-i==1)dp[i][j]=2;
                    else dp[i][j]=dp[i+1][j-1]+2;
                }
                for(int k=i;k<j;k++){
                    dp[i][j]=max(dp[i][j],dp[i][k]+dp[k+1][j]);
                }
            }
        }
        cout<<dp[0][len-1]<<endl;
    }
}