Parentheses Matrix (HDU6400)

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给定一个括号矩阵的宽度和高度，目标是构造一个具有最大好度的矩阵，好度定义为平衡的行和列的数量。题目提供了一种在n和m至少一个为奇数时的构造方法，以及在n和m都是偶数时的特殊情况的解决方案。

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Parentheses Matrix

Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 131072/131072 K (Java/Others)
Total Submission(s): 1418 Accepted Submission(s): 524
Special Judge

Problem Description

A parentheses matrix is a matrix where every element is either '(' or ')'. We define the goodness of a parentheses matrix as the number of balanced rows (from left to right) and columns (from up to down). Note that:

- an empty sequence is balanced;
- if A is balanced, then (A) is also balanced;
- if A and B are balanced, then AB is also balanced.

For example, the following parentheses matrix is a 2×4 matrix with goodness 3, because the second row, the second column and the fourth column are balanced:

)()(
()()

Now, give you the width and the height of the matrix, please construct a parentheses matrix with maximum goodness.

Input

The first line of input is a single integer T (1≤T≤50), the number of test cases.

Each test case is a single line of two integers h,w (1≤h,w≤200), the height and the width of the matrix, respectively.

Output

For each test case, display h lines, denoting the parentheses matrix you construct. Each line should contain exactly w characters, and each character should be either '(' or ')'. If multiple solutions exist, you may print any of them.

Sample Input

1 1

2 2

2 3

Sample Output

(

()

)(

(((

)))

Source

2018 Multi-University Training Contest 8

当 n 和 m 中至少有一个是奇数时，构造的方法是显然的。下面仅讨论n 和 m 都是偶数的情况。

当 n = 2 或 m = 2 的时候匹配数为 n + m - 2

当 n = 4 或 m = 4 的时候匹配数为 n + m - 3

其余情况匹配数为 n + m - 4

n = 4的时候可以这么构造

((((((

)))(((

((()))

))))))

m = 4 的时候可以这么构造

()()

(())

其余情况时，可以如下构造：

（保证左边界和上边界是左括号，右边界和下边界是右括号）

((((((((

()()()()

(()()())

()()()()

(()()())

))))))))
匹配数是 n+m−4。

#include <iostream>
using namespace std;
typedef long long ll;

int main()
{
    int T,n,m;
    cin>>T;
    while(T--)
    {
        cin>>n>>m;
        if(n%2==1)
        {
            for(int i=1;i<=n;i++)
            {
                for (int j=1;j<=m;j++)
                {
                    if (j%2==1)
                        cout<<"(";
                    else
                        cout<<")";
                }
                cout<<endl;
            }
        }
        else if(m%2==1)
        {
            for(int i=1;i<=n;i++)
            {
                for (int j=1;j<=m;j++)
                {
                    if (i%2==1)
                        cout<<"(";
                    else
                        cout<<")";
                }
                cout<<endl;
            }
        }
        else
        {
            if(n==2)
            {
                for(int j=1; j<=m; j++)
                    cout<<"(";
                cout<<endl;
                for(int j=1; j<=m; j++)
                    cout<<")";
                cout<<endl;
            }
            else if(n==4)
            {
                for(int j=1; j<=m; j++)
                    cout<<"(";
                cout<<endl;
                for(int j=1; j<=m; j++)
                {
                    if(j<=m/2)
                        cout<<")";
                    else
                        cout<<"(";
                }
                cout<<endl;
                for(int j=1; j<=m; j++)
                {
                    if(j>m/2)
                        cout<<")";
                    else
                        cout<<"(";
                }
                cout<<endl;
                for(int j=1; j<=m; j++)
                    cout<<")";
                cout<<endl;
            }
            else if(m==2)
            {
                for(int i=1; i<=n; i++)
                    cout<<"()"<<endl;
            }
            else if(m==4)
            {
                for(int i=1; i<=n; i++)
                {
                    if(i<=n/2)
                        cout<<"()()"<<endl;
                    else
                        cout<<"(())"<<endl;
                }
            }

            else
            {
                for(int j=1; j<=m; j++)
                {
                    cout<<"(";
                }
                cout<<endl;
                for(int i=2; i<=n-1; i++)
                {
                    if(i%2==1)
                    {
                        for(int j=1; j<=m; j++)
                        {
                            if(j%2==1)
                                cout<<"(";
                            else
                                cout<<")";
                        }
                    }
                    else
                    {
                        for(int j=1; j<=m; j++)
                        {
                            if(j==1)
                                cout<<"(";
                            else if(j==m)
                                cout<<")";
                            else
                            {
                                if(j%2==0)
                                    cout<<"(";
                                else
                                    cout<<")";
                            }
                        }
                    }
                    cout<<endl;
                }
                for(int j=1; j<=m; j++)
                {
                    cout<<")";
                }
                cout<<endl;
            }
        }
    }
    return 0;
}