POJ 2955 Brackets （DP Or 记忆化搜索 总结）

POJ - 2955 Brackets 

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

这个好像比那个here更简单一些

题意：求出互相匹配的括号的总数

思路：一道区间DP，dp[i][j]存的是i~j区间内匹配最大个数

网上多数的递推表达式不太懂都是这样的：

if(s[i]=='('&&s[j]==')'||s[i]=='['&&s[j]==']')
     dp[i][j]=dp[i+1][j-1]+2;
dp[i][j]=max{dp[i][k]+dp[k+1][j]};

还是我自己写的好想一些：

 dp[i][j]=max(dp[i][j-1], 2+dp[i][k-1]+dp[k+1][j-1])   i<=k<j    dp[i][j-1]为和谁都不匹配   2+dp[i][k-1]+dp[k+1][j-1]为和k匹配

不过好像每次看i好一点，我是看的j所在的匹配情况，不过大同小异

#include<iostream>
#include<cstdio>
#include<algorithm>
#include<cstring>
using namespace std;
const int N=105;
char s[N];
int dp[N][N];
inline bool check(int i,int j){
    return (s[i]=='['&&s[j]==']' || s[i]=='('&&s[j]==')');
}

int main(){
    while(~scanf("%s",s+1)){
        if(s[1]=='e') break;
        memset(dp,0,sizeof(dp));
        int n=strlen(s+1); ///还能这么测长度，长见识了
        for(int i=n;i>=1;i--)
            for(int j=i+1;j<=n;j++){
                dp[i][j]=dp[i][j-1]; ///和谁都不匹配
                for(int k=i;k<j;k++) ///出现匹配
                  if(check(k,j))
                    dp[i][j]=max(dp[i][j],2+dp[i][k-1]+dp[k+1][j-1]);
            }
        printf("%d\n",dp[1][n]);
    }
}

记忆化搜索方式

突然从这道题目里悟出区间的这种DP好想比之前的那种一般的DP还是和用记忆化搜索啊，因为这里涉及到区间了，所以划分成多个重复子问题更适合。

暂且这么理解吧

#include <iostream>
#include <string.h>
#include <stdio.h>
#include <algorithm>
using namespace std;
const int MAXN=110;
char str[MAXN];
int dp[MAXN][MAXN];

bool check(int i,int j){
  return (str[i]=='('&&str[j]==')')||(str[i]=='['&&str[j]==']');
}

int solve(int i,int j)
{
    if(dp[i][j]!=-1) return dp[i][j];
    if(j<=i)  return dp[i][j]=0;
    
    dp[i][j]=solve(i+1,j); ///和谁都不匹配情况
    for(int k=i+1;k<=j;k++)   ///和其中一个匹配
        if(check(i,k))
            dp[i][j]=max(dp[i][j],2+solve(i+1,k-1)+solve(k+1,j));
    return dp[i][j];
}

int main()
{
    while(scanf("%s",str)==1)
    {
        if(strcmp(str,"end")==0)break;
        memset(dp,-1,sizeof(dp));
        int n=strlen(str);
        printf("%d\n",solve(0,n-1));
    }
    return 0;
}