POJ2955 Brackets —— 区间DP

题目链接：https://vjudge.net/problem/POJ-2955

Brackets

Time Limit: 1000MS		Memory Limit: 65536K
Total Submissions: 9630		Accepted: 5131

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

题解：

求最多有多少对括号匹配。典型的区间dp。

方法一：

1.如果区间[l,r]的两端匹配，则左右各缩进一格，从而转化成处理[l+1, r-1]的区间。

2.不管是否符合条件1，都尝试去枚举分割点，使得整个区间分成两半，这样就可以把大区间的处理转化成两个小区间的处理。

记忆化搜索：

 #include <iostream>
 #include <cstdio>
 #include <cstring>
 #include <algorithm>
 #include <vector>
 #include <cmath>
 #include <queue>
 #include <stack>
 #include <map>
 #include <string>
 #include <set>
 using namespace std;
 typedef long long LL;
 const int INF = 2e9;
 const LL LNF = 9e18;
 const int MOD = 1e9+;
 const int MAXN = +;

 char s[MAXN];
 int dp[MAXN][MAXN];

 int dfs(int l, int r)
 {
     if(r<=l) return ;
     if(dp[l][r]!=-) return dp[l][r];

     if( (s[l]=='('&&s[r]==')')||(s[l]=='['&&s[r]==']') )    //如果两端匹配，则可以缩减范围
         dp[l][r] = dfs(l+, r-) + ;
     for(int k = l; k<r; k++)    //枚举分割点，分成两半
         dp[l][r] = max(dp[l][r], dfs(l, k)+dfs(k+, r));

     return dp[l][r];
 }

 int main()
 {
     while(scanf("%s", s+) && strcmp(s+, "end"))
     {
         memset(dp, -, sizeof(dp));
         cout<< dfs(, strlen(s+))* <<endl;
     }
 }

递推：

 #include <iostream>
 #include <cstdio>
 #include <cstring>
 #include <algorithm>
 #include <vector>
 #include <cmath>
 #include <queue>
 #include <stack>
 #include <map>
 #include <string>
 #include <set>
 using namespace std;
 typedef long long LL;
 const int INF = 2e9;
 const LL LNF = 9e18;
 const int MOD = 1e9+;
 const int MAXN = +;

 char s[MAXN];
 int dp[MAXN][MAXN];

 int main()
 {
     while(scanf("%s", s+) && strcmp(s+, "end"))
     {
         memset(dp, , sizeof(dp));
         int n = strlen(s+);
         for(int len = ; len<=n; len++)
         {
             for(int l = ; l<=n-len+; l++)
             {
                 int r = l+len-;
                 if( (s[l]=='('&&s[r]==')') || (s[l]=='['&&s[r]==']') )
                     dp[l][r] = dp[l+][r-] + ;
                 for(int k = l; k<r; k++)
                     dp[l][r] = max(dp[l][r], dp[l][k]+dp[k+][r]);
             }
         }
         printf("%d\n", dp[][n]*);
     }
     return ;
 }

方法二：

1.可知一个符号最多只能与一个符号匹配，那么对于当前的符号，我们就枚举其他符号与其匹配（不管是能匹配成功）。

2.假设区间为 [l, r]，为l枚举匹配符号，当枚举到k位置时，就把区间分割成了两部分：[l+1, k-1] 和 [k+1, r] 。从而就把大区间的求解转化为小区间的求解。

记忆化搜索：

 #include <iostream>
 #include <cstdio>
 #include <cstring>
 #include <algorithm>
 #include <vector>
 #include <cmath>
 #include <queue>
 #include <stack>
 #include <map>
 #include <string>
 #include <set>
 using namespace std;
 typedef long long LL;
 const int INF = 2e9;
 const LL LNF = 9e18;
 const int MOD = 1e9+;
 const int MAXN = +;

 char s[MAXN];
 int dp[MAXN][MAXN];

 int dfs(int l, int r)
 {
     if(r<=l) return ;
     if(dp[l][r]!=-) return dp[l][r];

     dp[l][r] = dfs(l+, r);
     for(int k = l+; k<=r; k++)
     {
         int ismatch = (s[l]=='('&&s[k]==')')||(s[l]=='['&&s[k]==']');
         int tmp =  dfs(l+, k-)+dfs(k+, r)+ismatch;
         dp[l][r] = max(dp[l][r], tmp);
     }
     return dp[l][r];
 }

 int main()
 {
     while(scanf("%s", s+) && strcmp(s+, "end"))
     {
         memset(dp, -, sizeof(dp));
         cout<< dfs(, strlen(s+))* <<endl;
     }
 }

递推：

 #include <iostream>
 #include <cstdio>
 #include <cstring>
 #include <algorithm>
 #include <vector>
 #include <cmath>
 #include <queue>
 #include <stack>
 #include <map>
 #include <string>
 #include <set>
 using namespace std;
 typedef long long LL;
 const int INF = 2e9;
 const LL LNF = 9e18;
 const int MOD = 1e9+;
 const int MAXN = +;

 char s[MAXN];
 int dp[MAXN][MAXN];

 int main()
 {
     while(scanf("%s", s+) && strcmp(s+, "end"))
     {
         memset(dp, , sizeof(dp));
         int n = strlen(s+);
         for(int len = ; len<=n; len++)
         {
             for(int l = ; l<=n-len+; l++)
             {
                 int r = l+len-;
                 dp[l][r] = dp[l+][r];
                 for(int k = l+; k<=r; k++)
                 {
                     int ismatch = (s[l]=='('&&s[k]==')')||(s[l]=='['&&s[k]==']');
                     dp[l][r] = max(dp[l][r], dp[l+][k-]+dp[k+][r]+ismatch);
                 }
             }
         }
         printf("%d\n", dp[][n]*);
     }
     return ;
 }