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+7;
const int MAXN = 100+10;


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

int dfs(int l, int r)
{
    if(r<=l) return 0;
    if(dp[l][r]!=-1) 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+1, r-1) + 1;
    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+1, r));

    return dp[l][r];
}

int main()
{
    while(scanf("%s", s+1) && strcmp(s+1, "end"))
    {
        memset(dp, -1, sizeof(dp));
        cout<< dfs(1, strlen(s+1))*2 <<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+7;
const int MAXN = 100+10;

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

int main()
{
    while(scanf("%s", s+1) && strcmp(s+1, "end"))
    {
        memset(dp, 0, sizeof(dp));
        int n = strlen(s+1);
        for(int len = 2; len<=n; len++)
        {
            for(int l = 1; l<=n-len+1; l++)
            {
                int r = l+len-1;
                if( (s[l]=='('&&s[r]==')') || (s[l]=='['&&s[r]==']') )
                    dp[l][r] = dp[l+1][r-1] + 1;
                for(int k = l; k<r; k++)
                    dp[l][r] = max(dp[l][r], dp[l][k]+dp[k+1][r]);
            }
        }
        printf("%d\n", dp[1][n]*2);
    }
    return 0;
}

 

 

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+7;
const int MAXN = 100+10;

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

int dfs(int l, int r)
{
    if(r<=l) return 0;
    if(dp[l][r]!=-1) return dp[l][r];

    dp[l][r] = dfs(l+1, r);
    for(int k = l+1; k<=r; k++)
    {
        int ismatch = (s[l]=='('&&s[k]==')')||(s[l]=='['&&s[k]==']');
        int tmp =  dfs(l+1, k-1)+dfs(k+1, r)+ismatch;
        dp[l][r] = max(dp[l][r], tmp);
    }
    return dp[l][r];
}

int main()
{
    while(scanf("%s", s+1) && strcmp(s+1, "end"))
    {
        memset(dp, -1, sizeof(dp));
        cout<< dfs(1, strlen(s+1))*2 <<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+7;
const int MAXN = 100+10;

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

int main()
{
    while(scanf("%s", s+1) && strcmp(s+1, "end"))
    {
        memset(dp, 0, sizeof(dp));
        int n = strlen(s+1);
        for(int len = 2; len<=n; len++)
        {
            for(int l = 1; l<=n-len+1; l++)
            {
                int r = l+len-1;
                dp[l][r] = dp[l+1][r];
                for(int k = l+1; k<=r; k++)
                {
                    int ismatch = (s[l]=='('&&s[k]==')')||(s[l]=='['&&s[k]==']');
                    dp[l][r] = max(dp[l][r], dp[l+1][k-1]+dp[k+1][r]+ismatch);
                }
            }
        }
        printf("%d\n", dp[1][n]*2);
    }
    return 0;
}