HDU

Parentheses Matrix

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 131072/131072 K (Java/Others)
Total Submission(s): 0    Accepted Submission(s): 0
Special Judge


Problem Description
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 (1T50), the number of test cases.

Each test case is a single line of two integers h,w (1h,w200), 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
(
()
)(
(((
)))
 
 
 
构造题。找出每行每列最大匹配数的矩阵。
首先分奇偶性讨论。
1.奇奇情况为0,任意输出。
2.奇偶情况最大匹配为偶数值。
3.偶偶要分两种情况,最大匹配为max(h,w)+min(h,w)/2-1或h+w-4,
行和列较小的值若小于等于6,第一行(列)为“(”,最后一行(列)为“)”,大于6则第一行列全为“(”,最后一行列全为“)”。
其他的位置行列下标和若偶为“(”,奇为“)”。
 
 
#include <bits/stdc++.h>
using namespace std;
const int INF = 0x3f3f3f3f;
const int MAX = 205;
typedef long long LL;

int main(void)
{
    int t,n,m,i,j;
    scanf("%d",&t);
    while(t--){
        scanf("%d%d",&n,&m);
        if((n&1)&&(m&1)){
            for(i=1;i<=n;i++){
                for(j=1;j<=m;j++){
                    printf("(");
                }
                printf("
");
            }
        }
        else if(n&1){
            for(i=1;i<=n;i++){
                for(j=1;j<=m;j++){
                    if(j&1) printf("(");
                    else printf(")");
                }
                printf("
");
            }
        }
        else if(m&1){
            for(i=1;i<=n;i++){
                for(j=1;j<=m;j++){
                    if(i&1) printf("(");
                    else printf(")");
                }
                printf("
");
            }
        }
        else{
            if(n>=m){
                for(i=1;i<=n;i++){
                    for(j=1;j<=m;j++){
                        if(i==1&&(m/2-1)>2){
                            printf("(");
                            continue;
                        }
                        if(i==n&&(m/2-1)>2){
                            printf(")");
                            continue;
                        }
                        if(j==1){
                            printf("(");
                            continue;
                        }
                        if(j==m){
                            printf(")");
                            continue;
                        }
                        if((i+j)&1){
                            printf(")");
                        }
                        else{
                            printf("(");
                        }
                    }
                    printf("
");
                }
            }
            else{
                for(i=1;i<=n;i++){
                    for(j=1;j<=m;j++){
                        if(j==1&&(n/2-1)>2){
                            printf("(");
                            continue;
                        }
                        if(j==m&&(n/2-1)>2){
                            printf(")");
                            continue;
                        }
                        if(i==1){
                            printf("(");
                            continue;
                        }
                        if(i==n){
                            printf(")");
                            continue;
                        }
                        if((i+j)&1){
                            printf(")");
                        }
                        else{
                            printf("(");
                        }
                    }
                    printf("
");
                }
            }
        }
    }
    return 0;
}
原文地址:https://www.cnblogs.com/yzm10/p/9482992.html