UVA 10267

 Problem E: Graphical Editor 

Problem

The simple graphical editor deals with a rectangular table M×N (1<=M,N<=250). Each pixel of the table has its colour. The picture is formed from this square pixels.

The problem is to write a program, which simulates an interactive work of the graphical editor.

Input

Input consists of the editor commands, one per line. Each command is represented by one Latin capital placed in the very beginning of the line. If the command supposes parameters, all the parameters will be given in the same line separated by space. As the parameters there may be: the coordinates of the pixel - two integers, the first one is the column number and belongs to 1..M, the second one is the row number and belongs to 1..N, the origin is in the upper left corner of the table; the colour - the Latin capital; file name - in MSDOS 8.3 format.

The editor deals with the following commands:

I M NCreates a new table M×N. All the pixels are colored in white (O).CClears the table. The size remains the same. All the pixels became white (O).L X Y CColors the pixel with coordinates (X,Y) in colour C.V X Y1 Y2 CDraws the vertical segment in the column X between the rows Y1 and Y2 inclusive in colour C.H X1 X2 Y CDraws the horizontal segment in the row Y between the columns X1 and X2 inclusive in colour C.K X1 Y1 X2 Y2 CDraws the filled rectangle in colour C. (X1,Y1) is the upper left corner, (X2,Y2) is the lower right corner of the rectangle.F X Y CFills the region with the colour C. The region R to be filled is defined as follows. The pixel (X,Y) belongs to this region. The other pixel belongs to the region R if and only if it has the same colour as pixel (X,Y) and a common side with any pixel which belongs to this region.S NameWrites the picture in the file Name.XTerminates the session.

Output

Every time the command S NAME meets, you should output the file name NAME and the current table, row by row. Each row is represented by a pixels' colours series, see the output sample.

Errors

If as a command there will be a character different from I, C, L, V, H, K, F, S, X, the editor should ignore the whole line and pass to the next command.

In case of other errors the program behaviour is unpredictable.

Sample Input

I 5 6L 2 3 AS one.bmpG 2 3 JF 3 3 JV 2 3 4 WH 3 4 2 ZS two.bmpX

Sample Output

one.bmpOOOOOOOOOOOAOOOOOOOOOOOOOOOOOOtwo.bmpJJJJJJJZZJJWJJJJWJJJJJJJJJJJJJ

这道题是一道极好的小模拟 ， 至少我写得很爽 ， 虽然WA了两次

主要trick有一下几个 ：首先x1 ， x2 --- 不确定大小关系；其次这题用Y表示横坐标，X表示纵坐标 ，我就WA在这儿了。

      其中还有个小递归 ， 要想一下 ，其他没啥了

#include <iostream>#include <cstdio>#include <cstring>#include <algorithm>char a[255][255];char G;char s[20];int vis[255][255];int m , n;void find(int  i , int j , char G){    if(vis[i][j]||i < 1 || j < 1||i > m ||j > n)return ;    vis[i][j] = 1;    if(a[i - 1][j] == a[i][j])find(i - 1 , j , G);    if(a[i + 1][j] == a[i][j])find(i + 1 , j , G);    if(a[i][j - 1] == a[i][j])find(i , j - 1 , G);    if(a[i][j + 1] == a[i][j])find(i , j + 1 , G);    a[i][j] = G;}using namespace std;int main(){    char c ;    int x1 , y1 , x2 , y2 , X , Y;    //freopen("in.txt" , "r" , stdin);    while(~scanf("%c",&c))    {        if(c == 'X')break;        switch(c)        {            case 'I':                cin >> n >> m;                for(int i = 1 ; i <= m ; i++)                {                    for(int j = 1 ; j <= n ; j++)                    {                        a[i][j] = 'O';                    }                }                break;            case 'C':                for(int i = 1 ; i <= m ; i++)                {                    for(int j = 1 ; j <= n ; j++)                    {                        a[i][j] = 'O';                    }                }                break;            case 'L':                cin >> y1 >>  x1 >> G;                a[x1][y1] = G;                break;            case 'V':                cin >> Y >> x1 >> x2 >> G;                if(x1 > x2)swap(x1 , x2);                for(int i = x1 ; i <= x2 ; i++)                {                    a[i][Y] = G;                }                break;            case 'H':                cin >> y1 >> y2 >> X >> G;                if(y1 > y2)swap(y1 , y2);                for(int i = y1 ; i <= y2 ; i++)                {                    a[X][i] = G;                }                break;            case 'K':                cin >> y1 >> x1 >> y2 >> x2 >> G;                if(x1 > x2)swap(x1 , x2);                if(y1 > y2)swap(y1 , y2);                for(int i = x1 ; i <= x2 ; i++)                {                    for(int j = y1 ; j <= y2 ; j++)                    {                        a[i][j] = G;                    }                }                break;            case 'F':                cin >> Y >> X >> G;                memset(vis , 0 , sizeof(vis));                find(X , Y , G);                break;            case 'S':                cin >> s;                cout << s << endl;                for(int i = 1 ; i <= m ; i ++)                {                    for(int j = 1  ; j <= n ; j++)                    {                        cout << a[i][j];                    }                    cout << endl;                }                break;            default :                continue;        }    }    return 0;}