Parentheses Matrix
Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 131072/131072 K (Java/Others)
Total Submission(s): 397 Accepted Submission(s): 143
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
思路：
这里n和m可以互换》》》
1. 对于行列都是奇数的情况，肯定是一个没有
2.对于行列中有一个是奇数一个是偶数的情况，答案肯定是奇数
3. 对于行列都是偶数的情况我们这样考虑，如果我不放弃第一列，也就是使得第一列为匹配的，所以（1,1）肯定是左括号，如果第一列是匹配并且n>=6 ，那么就相当于我们放弃了n/2行， 并且，如果我们使得第一行为匹配的呢还是最后一列是匹配的呢。这两个也是一个矛盾，所以在n>=6 的情况下其实是放弃第一列，最优的状态就是n+m-4,
((((((((
()()()()
(()()())
()()()()
(()()())
))))))))
在n<=4 的状态下，如果我们考虑不放弃第一列，那么我们不放弃第一列，那么肯定会有n/2行是不匹配的，并且右上角如何选择会减掉一个，所以最优的状态其实是n+m-3 .
((((((
)))(((
((()))
))))))
代码： 上半部分垃圾暴力，下半部分代码
/*#include<bits/stdc++.h>
using namespace std;
const int N =15;
int mp[N][N];
int n,m;
int maxx=-1;
void hang(int &sum)
{
stack<char >s;
for(int i=1;i<=n;i++){
while(!s.empty()) s.pop();
int f=0;
for(int j=1;j<=m;j++){
if(mp[i][j]==1){
s.push('(');
}
else{
if(s.empty()){
f=1; break;
}
else{
s.pop();
}
}
}
if(f||!s.empty()) continue;
sum++;
}
}
void lie(int &sum)
{
stack< char >s;
for(int j=1;j<=m;j++){
while(!s.empty()) s.pop();
int f=0;
for(int i=1;i<=n;i++){
if(mp[i][j]==1){
s.push('(');
}
else{
if(s.empty()){
f=1; break;
}
else s.pop();
}
}
if(f||!s.empty()) continue;
sum++;
}
}
void jud()
{
int tot=0;
hang(tot);
lie(tot);
if(tot>maxx){
maxx=tot;
cout<<"maxx "<<maxx<<endl;
for(int i=1;i<=n;i++){
for(int j=1;j<=m;j++){
if(mp[i][j]==1){
printf("( ");
}
else printf(") ");
}
printf("\n");
}
}
}
void dfs(int x,int y)
{
if(x==n+1){
jud();
return ;
}
mp[x][y]=1;
int tmpx,tmpy;
if(y==m){
tmpx=x+1; tmpy=1;
}
else{
tmpx=x; tmpy=y+1;
}
dfs(tmpx,tmpy);
mp[x][y]=2;
dfs(tmpx,tmpy);
}
int main()
{
while(cin>>n>>m)
{
maxx=-1;
dfs(1,1);
}
return 0;
}
*/
#include<bits/stdc++.h>
using namespace std;
const int N =205;
int n,m;
int mp[N][N];
int main()
{
//freopen("d.in","r",stdin);
//freopen("dd.out","w",stdout);
int T;
scanf("%d",&T);
while(T--)
{
scanf("%d %d",&n,&m);
int odd=0,even=0;
if(n%2==1) odd++;
else even++;
if(m%2==1) odd++;
else even++;
if(odd==2)
{
for(int i=1; i<=n; i++)
{
for(int j=1; j<=m; j++) mp[i][j]=1;
}
}
else if(even&&odd)
{
if(n%2==0) /// 如果行是偶数
{
for(int i=1; i<=n/2; i++)
{
for(int j=1; j<=m; j++)
{
mp[i][j]=1;
}
}
for(int i=n/2+1; i<=n; i++)
{
for(int j=1; j<=m; j++)
{
mp[i][j]=2;
}
}
}
else
{
for(int i=1; i<=n; i++)
{
for(int j=1; j<=m; j++)
{
if(j<=m/2) mp[i][j]=1;
else mp[i][j]=2;
}
}
}
}
else
{
if(n<=m)
{
for(int i=1; i<=n/2; i++)
{
if(i%2==1)
{
for(int j=1; j<=m; j++) mp[i][j]=1;
}
else
{
for(int j=1; j<=m; j++)
{
if(j<=m/2) mp[i][j]=1;
else mp[i][j]=2;
}
}
}
int tot=0;
for(int i=n; i>n/2; i--)
{
++tot;
if(tot%2==1)
{
for(int j=1; j<=m; j++) mp[i][j]=2;
}
else
{
for(int j=1; j<=m; j++)
{
if(j<=m/2) mp[i][j]=2;
else mp[i][j]=1;
}
}
}
}
else
{
for(int i=1; i<=n/2; i++)
{
for(int j=1; j<=m; j++)
{
if(j%2==1) mp[i][j]=1;
else mp[i][j]=2;
}
}
for(int i=n/2+1; i<=n; i++)
{
for(int j=1; j<=m; j++)
{
if(j<=m/2) mp[i][j]=1;
else mp[i][j]=2;
}
}
}
if(n>=6&&m>=6)
{
for(int j=1; j<=m; j++) mp[1][j]=1;
for(int i=2; i<=n-1; i++)
{
if(i%2==0)
{
mp[i][1]=1;
mp[i][m]=2;
for(int j=2; j<m; j++)
{
if(j%2==0) mp[i][j]=1;
else mp[i][j]=2;
}
}
else
{
for(int j=1; j<=m; j++)
{
if(j%2==1) mp[i][j]=1;
else mp[i][j]=2;
}
}
}
for(int j=1; j<=m; j++) mp[n][j]=2;
}
}
for(int i=1; i<=n; i++)
{
for(int j=1; j<=m; j++)
{
if(mp[i][j]==1) printf("(");
else printf(")");
}
printf("\n");
}
}
return 0;
}
/*
((((((
)))(((
((()))
))))))
((((((((
()()()()
(()()())
()()()()
(()()())
))))))))
*/