Parentheses Matrix
Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 131072/131072 K (Java/Others)
Total Submission(s): 973 Accepted Submission(s): 385
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
3 1 1 2 2 2 3
Sample Output
( () )( ((( )))
Source
2018 Multi-University Training Contest 8
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#include<bits/stdc++.h>
using namespace std;
#define debug puts("YES");
#define rep(x,y,z) for(int (x)=(y);(x)<(z);(x)++)
#define read(x,y) scanf("%d%d",&x,&y)
/*
给定规格要求构造规格中的一个括号矩阵，
每行每列能够构造出的规范括号序列数量最多。
没啥其他高大上的推导，
感觉就是打表找规律，
唯一麻烦的情况就是n和m均为偶数的情况，
当规格中有一个是2的时候，也很好观察，
最大的数量是max(n,m)，
但当一个是4的时候，就需要打表了。。。
因为最大值是n+m-3，因为4的那个规模，其中可以多凑出来一个括号序列，
((((((
)))(((
((()))///多出来的一个序列
))))))
其余情况时，最大值是n+m-4.
((((((
奇数：()()()
偶数：(()())
))))))
牺牲了第一行和第一列，最后一行和最后一列。
*/
const int maxn =2e5+5;
int n,m,table[250][250];
void draw(int n,int m,int v)
{
if(v==0) rep(i,0,n) rep(j,0,m) table[i][j]=1;
else if(v==1||v==3) rep(i,0,n) for(int j=0;j<m;j+=2) table[i][j]=1,table[i][j+1]=-1;
else if(v==2||v==4) rep(i,0,m) for(int j=0;j<n;j+=2) table[j][i]=1,table[j+1][i]=-1;
}
int main()
{
int t;scanf("%d",&t);
while(t--)
{
scanf("%d%d",&n,&m);
if(n&1)
{
if(m&1) draw(n,m,0);/// 无法构成
else draw(n,m,1);
}
else
{
if(m&1) draw(n,m,2);
else
{///最复杂的部分
///n和m中有2
if(min(n,m)==2)
{
if(n==2) for(int i=0;i<m;i++) table[0][i]=1,table[1][i]=-1;
else for(int i=0;i<n;i++) table[i][0]=1,table[i][1]=-1;
}
else if(min(n,m)==4)
{
///debug
if(n==4)
{
for(int i=0;i<m;i++) table[0][i]=1,table[3][i]=-1;
for(int i=0;i<m;i++)
{
if(i<m/2) table[1][i]=-1,table[2][i]=1;
else table[1][i]=1,table[2][i]=-1;
}
}
else if(m==4)
{
for(int i=0;i<n;i++) table[i][0]=1,table[i][3]=-1;
for(int i=0;i<n;i++)
{
if(i<n/2) table[i][1]=-1,table[i][2]=1;
else table[i][1]=1,table[i][2]=-1;
}
}
}
else
{
for(int i=0;i<m;i++) table[0][i]=1,table[n-1][i]=-1;
for(int i=0;i<m;i+=2) table[1][i]=1,table[1][i+1]=-1;
table[2][0]=1,table[2][m-1]=-1;
for(int i=1;i<m-1;i+=2) table[2][i]=1,table[2][i+1]=-1;
for(int i=1;i<n-1;i++)
{
if(i&1) memcpy(table[i],table[1],sizeof(table[1]));
else memcpy(table[i],table[2],sizeof(table[2]));
}
}
}
}
for(int i=0;i<n;i++,puts("")) for(int j=0;j<m;j++) printf("%c",(table[i][j]==1)?'(':')');
}
return 0;
}