烟大 Contest1024 - 《挑战编程》第一章：入门 Problem E: Graphical Editor（模拟控制台命令形式修改图形）

Problem E: Graphical Editor

Time Limit: 1 Sec  Memory Limit: 64 MB
Submit: 2  Solved: 2
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Description

Graphical editors such as Photoshop allow us to alter bit-mapped images in the same way that text editors allow us to modify documents. Images are represented as an M x N array of pixels, where each pixel has a given color. Your task is to write a program which simulates a simple interactive graphical editor.

Input

The input consists of a sequence of editor commands, one per line. Each command is represented by one capital letter placed as the first character of the line. If the command needs parameters, they will be given on the same line separated by spaces. Pixel coordinates are represented by two integers, a column number between 1...M and a row number between 1...N, where 1M, N250. The origin sits in the upper-left corner of the table. Colors are specified by capital letters.

I M N	Create a new M x N image with all pixels initially colored white (O).
C	Clear the table by setting all pixels white (O). The size remains unchanged.
L X Y C	Colors the pixel (X, Y) in color (C).
V X Y1 Y2 C	Draw a vertical segment of color (C) in columnX, between the rows Y1 and Y2 inclusive.
H X1 X2 Y C	Draw a horizontal segment of color (C) in the row Y, between the columns X1 and X2inclusive.
K X1 Y1 X2 Y2 C	Draw a filled rectangle of color C, where (X1,Y1) is the upper-left and (X2, Y2) the lower right corner.
F X Y C	Fill the region R with the color C, where R is defined as follows. Pixel (X, Y) belongs to R. Any other pixel which is the same color as pixel(X, Y) and shares a common side with any pixel in R also belongs to this region.
S Name	Write the file name in MSDOS 8.3 format followed by the contents of the current image.
X	Terminate the session.

Output

On every command S NAME, print the filename NAME and contents of the current image. Each row is represented by the color contents of each pixel. See the sample output. Ignore the entire line of any command defined by a character other than I, C, L, V, H, K, F, S, or X, and pass on to the next command. In case of other errors, the program behavior is unpredictable.

Sample Input

I 5 6

L 2 3 A

S one.bmp

G 2 3 J

F 3 3 J

V 2 3 4 W

H 3 4 2 Z

S two.bmp

X

Sample Output

one.bmp

OOOOO

OOOOO

OAOOO

OOOOO

OOOOO

OOOOO

two.bmp

JJJJJ

JJZZJ

JWJJJ

JWJJJ

JJJJJ

JJJJJ

HINT

---

　　模拟题。模拟的是控制台输命令的形式操作图片。

　　用switch()语句分别定义每一个命令。

　　需要注意的是F（填充命令），它的意思是将(X,Y)点的像素颜色相同的区域全部填充为C颜色。

　　其中样例输入里面的G命令是错误命令，意思是提示你遇到错误命令的时候不做处理。

　　另外，S NAME命令里提到了 MSDOS 8.3短文件名 格式输出，它的意思是：

　　　　8代表主文件名长度不超过8个字符。

　　　　3代表后缀名长度不超过3个字符。

　　　　且文件名内不能包括空格。

My code：

 #include <iostream>
 #include <string.h>
 using namespace std;
 char a[][];
 int main()
 {
     //command
     char com;
     //C
     int M,N;
     //L
     int lx,ly;
     char lc;
     //V
     int vx,vy1,vy2;
     char vc;
     //H
     int hx1,hx2,hy;
     char hc;
     //K
     int kx1,kx2,ky1,ky2;
     char kc;
     //F
     int fx,fy;
     char fc,cc;
     //S
     string l;
     while(cin>>com){
         if(com=='X')    //遇到X退出
             break;
         if(com!='I' && com!='C' && com!='L' && com!='V' && com!='H' && com!='K' && com!='F' && com!='S'){    //其他命令退出
             getline(cin,l);
             continue;
         }
         switch(com){
         case 'I':
             cin>>M>>N;
             for(int i=;i<=N;i++)   //创建M*N的空白(O)画板
                 for(int j=;j<=M;j++){
                     a[i][j]='O';
                 }
             break;
             //默认全部为O
         case 'C':
             //清空所有色彩为O
             for(int i=;i<=N;i++)   //清空画板
                 for(int j=;j<=M;j++){
                     a[i][j]='O';
                 }
         case 'L':
             cin>>lx>>ly>>lc;
             a[ly][lx]=lc;   //将lx,ly位置的颜色填充为lc
             break;
         case 'V':
             cin>>vx>>vy1>>vy2>>vc;
             for(int i=vy1;i<=vy2;i++)
                 a[i][vx]=vc;    //将x列vy1到vy2的像素颜色填充为vc
             break;
         case 'H':
             cin>>hx1>>hx2>>hy>>hc;
             for(int i=hx1;i<=hx2;i++)
                 a[hy][i]=hc;    //将y行vx1到vx2的像素颜色填充为hc
             break;
         case 'K':
             cin>>kx1>>kx2>>ky1>>ky2>>kc;
             for(int i=ky1;i<=ky2;i++)   //填充kx1,kx2,ky1,ky2区域为kc颜色
                 for(int j=kx1;j<=kx2;j++){
                     a[i][j]=kc;
                 }
             break;
         case 'F':
             cin>>fx>>fy>>fc;
             cc=a[fy][fx];
             for(int i=;i<=N;i++)   //填充画板与x，y点颜色相同的区域颜色为C
                 for(int j=;j<=M;j++){
                     if(a[i][j]==cc)
                         a[i][j]=fc;
                 }
             break;
         case 'S':
             cin>>l;
             cout<<l<<endl;  //先输出文件名
             for(int i=;i<=N;i++){   //显示
                 for(int j=;j<=M;j++)
                     cout<<a[i][j];
                 cout<<endl;
             }
             break;
         }
     }
     return ;
 }

 /**************************************************************
     Problem: 1498
     User: freecode
     Language: C++
     Result: Accepted
     Time:0 ms
     Memory:1328 kb
 ****************************************************************/

Freecode : www.cnblogs.com/yym2013