Parentheses Matrix 【2018 Multi-University Training Contest 8 】

最新推荐文章于 2019-07-10 15:41:00 发布

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本文探讨了如何构造一个括号矩阵，使其行和列的平衡数量最大化。通过分析不同情况下的规律，提供了一种算法实现，适用于宽度和高度不超过200的矩阵。

http://acm.hdu.edu.cn/showproblem.php?pid=6400

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

3 1 1 2 2 2 3

Sample Output

( () )( ((( )))

题意：

尽量让行匹配的数目和列匹配的数目尽量大

分析：

自己画画然后找到规律，我就找到了n,m都为偶数n<=6的规律，后面的那种神操作我没有想到

代码：【虽然长，但是超级好理解】

#include<bits/stdc++.h>
#define ll long long
using namespace std;
char a[505][505];
int main()
{
    int t,i,j,n,m;
    scanf("%d",&t);
    while(t--)
    {
        scanf("%d%d",&n,&m);
        if(n%2&&m%2)
        {
            for(i=0;i<n;i++)
            {
                for(j=0;j<m;j++)
                    printf("(");
                printf("\n");
            }
            continue;
        }
        else
        if(n%2)
        {
            for(i=0;i<n;i++)
            {
                for(j=0;j<m;j++){
                    if(j%2)
                        a[i][j]=')';
                    else
                        a[i][j]='(';
                }
            }
        }
        else
        if(m%2)
        {
            for(i=0;i<m;i++)
            {
                for(j=0;j<n;j++){
                    if(j%2)
                        a[j][i]=')';
                    else
                        a[j][i]='(';
                }
            }
        }
        else
        {
            if(n<=6||m<=6){
            if(n>=m)
            {
                for(i=0;i<n;i++)
                {
                    if(i%2==1)
                    {
                        a[i][0]='(';
                        for(j=1;j<m-1;j++){
                            if(j%2==1)
                                a[i][j]='(';
                            else
                                a[i][j]=')';
                        }
                        a[i][j]=')';
                    }
                    else
                    {
                        for(j=0;j<m;j++){
                            if(j%2==0)
                                a[i][j]='(';
                            else
                                a[i][j]=')';
                        }
                    }
                }
            }
            else
            {
                for(i=0;i<m;i++)
                {
                    if(i%2==1)
                    {
                        int len;
                        a[0][i]='(';
                        len=n-1;
                        for(j=1;j<len;j++){
                            if(j%2==1)
                                a[j][i]='(';
                            else
                                a[j][i]=')';
                        }
                        a[j][i]=')';
                    }
                    else
                    {
                        for(j=0;j<n;j++){
                            if(j%2==0)
                                a[j][i]='(';
                            else
                                a[j][i]=')';
                        }
                    }
                }
            }
            }
            else
            {
                for(i=0;i<m;i++)
                    a[0][i]='(',a[n-1][i]=')';
                for(i=0;i<n;i++)
                    a[i][0]='(',a[i][m-1]=')';
                for(i=1;i<n-1;i++)
                {
                    if(i%2){
                        for(j=1;j<m-1;j++){
                            if(j%2)
                                a[i][j]=')';
                            else
                                a[i][j]='(';
                        }
                    }
                    else
                    {
                        for(j=1;j<m-1;j++){
                            if(j%2)
                                a[i][j]='(';
                            else
                                a[i][j]=')';
                        }
                    }
                }
            }
        }
        for(i=0;i<n;i++)
        {
            for(j=0;j<m;j++)
                printf("%c",a[i][j]);
            printf("\n");
        }
    }
    return 0;
}