旋转矩阵

题目描述

景驰公司自成立伊始，公司便将“推动智能交通的发展，让人类的出行更安全，更高效，更经济，更舒适”作为公司使命，通过产业融合、建设智能汽车出行行业的方式，打造“利国、利民、利公司、利个人”的无人驾驶出行系统。公司的愿景是成为中国第一、世界一流的智能出行公司。

有一天，景驰公司的工程师在真车上做测试。

景驰公司的试验车上面有一个奇怪的图案，这是一个n*m的矩阵，这辆车可以到处开，每次可以左旋右旋，小明想知道转完之后的图案是怎么样的

具体来说：有一个n*m的字符矩阵，只包含3种字符（‘+’‘-’，‘|’）,通过一通乱旋之后变成什么样子？

输入描述:

第一行测试样例数T（0<T<=100）
每个测试样例第一行两个正整数n,m(0<n,m<=30)
接下来的n行是一个n*m的字符矩阵
字符矩阵之后是一串只包含‘L’(左旋)和‘R’（右旋）的字符串，长度不超过1000
每个样例间输出一个空行

输出描述:

第一行两个正整数n,m
接下来的n行是一个n*m的字符矩阵
每个样例后面输出一个空行

示例1

输入

2
2 3
+-+
|+|
LLRRR

3 2
-+
+|
-+
LLL

输出

3 2
-+
+|
-+

2 3
|+|
+-+

备注:

左旋即逆时针旋转，右旋即顺时针旋转
-通过一次左旋或右旋会变成|
|通过一次左旋或右旋会变成-

方法：旋转后有四种类型，将每种类型都列出来

#include <stdio.h>

#include <math.h>

#include <iostream>

using namespace std;

int main()

{

    char ch[50][50] , wq[50][50];

    int n, m , T;

    scanf("%d" , &T);

    while(T--)

    {

        string s;

        int ans = 0 ;

        int a = 0 , b = -1;

        scanf("%d %d" , &n , &m);

        for(int i = 0 ; i < n ; i++)

        {

            b++;

            for(int j = 0 ; j < m ; j++)

            {

                cin >> ch[i][j];

                wq[a][b] = ch[i][j];

                a++;

            }

            a = 0;

        }

        cin >> s;

        for(int i = 0 ; i < s.length() ; i++)

        {

            if(s[i] == 'L')

            {

                ans--;

            }

            else if(s[i] == 'R')

            {

                ans++;

            }

        }

        if(ans < 0)

        {

            ans+=400;

            ans = ans %4;

             

        }

        else

        {

            ans = ans%4;

        }

ans = ans + '0';

//  printf("ans = %c\n" , ans);

//cin >> ans;

//ans += '0';

            switch(ans)

            {

                case '0':

                    printf("%d %d\n" , n , m);

                    for(int i = 0 ; i < n ; i++)

                    {

                        for(int j = 0 ; j < m ; j++)

                        {

                        cout << ch[i][j];

                        }

                        cout << endl;

                    }

                    break;

                        case '1':

                            printf("%d %d\n" , m , n);

                for(int i = 0 ; i < m; i++)

                {

                    for(int j = n-1 ; j >= 0; j--)

                    {

                            if(wq[i][j] == '|')

                            {

                                wq[i][j] = '-';

                            }

                            else if(wq[i][j] == '-')

                            {

                                wq[i][j] = '|';

                            }

                            cout << wq[i][j];

                    }

                    cout << endl;

                }

                break;

                    case '2':

                        printf("%d %d\n" , n , m);

                    for(int i = n-1 ; i >= 0 ; i--)

                    {

                        for(int j = m - 1 ; j >= 0; j--)

                        {

                        cout << ch[i][j];

                        }

                        cout << endl;

                    }

                    break;

                    case '3':

                        printf("%d %d\n" , m , n);

                for(int i = m-1 ; i >= 0; i--)

                {

                    for(int j = 0 ; j < n; j++)

                    {

                            if(wq[i][j] == '|')

                            {

                                wq[i][j] = '-';

                            }

                            else if(wq[i][j] == '-')

                            {

                                wq[i][j] = '|';

                            }

                            cout << wq[i][j];

                    }

                    cout << endl;

                }

            }

    cout << endl;    

    }

    return 0 ;

}