• 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;
    }
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  • 原文地址:https://www.cnblogs.com/yzm10/p/9482992.html
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