poj2955

maksyuki 发表于 oj 分类，标签: DP-区间DP

15 9月 2015

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 msuch that for indices i₁, i₂, …, i_m where 1 ≤ i₁ < i₂ < … < i_m ≤ n, a_i₁a_i₂ … a_i_m 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

Source

Stanford Local 2004

题目类型：区间DP

算法分析：使用数组dp[i][j]表示下标从ai~aj中匹配的括号的个数，然后对于每一个ak(i <= k <= j)，如果ai和ak是匹配的，则dp[i][j] = max (dp[i][j], dp[i+1][k-1] + dp[k+1][j] + 2)

否则dp[i][j] = max (dp[i][j], dp[i+1][j])，初始时dp[i][i] = 0，最后输出dp[0][len-1]，其中len表示字符串的长度

#include <set>
#include <bitset>
#include <list>
#include <map>
#include <stack>
#include <queue>
#include <deque>
#include <string>
#include <vector>
#include <ios>
#include <iostream>
#include <fstream>
#include <sstream>
#include <iomanip>
#include <algorithm>
#include <utility>
#include <complex>
#include <numeric>
#include <functional>
#include <cmath>
#include <ctime>
#include <climits>
#include <cstdarg>
#include <cstdio>
#include <cstdlib>
#include <cstring>
#include <cctype>
#include <cassert>

#define lson rt << 1, l, m
#define rson rt << 1 | 1, m + 1, r

using namespace std;

const int INF = 0x7FFFFFFF;
const double EPS = 1e-10;
const double PI = 2 * acos (0.0);
const int MOD = 7;
const int maxn = 500 + 66;

char ans[maxn];
int dp[maxn][maxn];

int main()
{
//    ifstream cin ("aaa.txt");

    while (cin >> ans)
    {
        if (!strcmp (ans, "end"))
            break;

        int len = strlen (ans);
        memset (dp, 0, sizeof (dp));

        for (int i = 2; i <= len; i++)
        {
            for (int j = 0; j + i - 1 < len; j++)
            {
                int temp = j + i - 1;
                for (int k = j; k <= temp; k++)
                {
                    if ((ans[j] == '(' && ans[k] == ')') || (ans[j] == '[' && ans[k] == ']'))
                        dp[j][temp] = max (dp[j][temp], dp[j+1][k-1] + dp[k+1][temp] + 2);

                    else
                        dp[j][temp] = max (dp[j][temp], dp[j+1][temp]);
                }
            }
        }


        cout << dp[0][len-1] << endl;
    }

return 0;
}

#include <set>

#include <bitset>

#include <list>

#include <map>

#include <stack>

#include <queue>

#include <deque>

#include <string>

#include <vector>

#include <ios>

#include <iostream>

#include <fstream>

#include <sstream>

#include <iomanip>

#include <algorithm>

#include <utility>

#include <complex>

#include <numeric>

#include <functional>

#include <cmath>

#include <ctime>

#include <climits>

#include <cstdarg>

#include <cstdio>

#include <cstdlib>

#include <cstring>

#include <cctype>

#include <cassert>

#define lson rt << 1, l, m

#define rson rt << 1 | 1, m + 1, r

using namespace std;

const int INF = 0x7FFFFFFF;

const double EPS = 1e-10;

const double PI = 2 * acos (0.0);

const int MOD = 7;

const int maxn = 500 + 66;

char ans[maxn];

int dp[maxn][maxn];

int main()

{

// ifstream cin ("aaa.txt");

while (cin >> ans)

{

if (!strcmp (ans, "end"))

break;

int len = strlen (ans);

memset (dp, 0, sizeof (dp));

for (int i = 2; i <= len; i++)

{

for (int j = 0; j + i - 1 < len; j++)

{

int temp = j + i - 1;

for (int k = j; k <= temp; k++)

{

if ((ans[j] == '(' && ans[k] == ')') || (ans[j] == '[' && ans[k] == ']'))

dp[j][temp] = max (dp[j][temp], dp[j+1][k-1] + dp[k+1][temp] + 2);

else

dp[j][temp] = max (dp[j][temp], dp[j+1][temp]);

}

cout << dp[0][len-1] << endl;

}

return 0;

}

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