PC/UVa 题号: 110105/10267 Graphical Editor (图形化编辑器)题解
#include<cstdio>
#include<iostream>
#include<string>
#include<algorithm>
#include<iterator>
#include<cstring>
#include<set>
using namespace std;
int m;
int n;
//bmp[y][x]
char bmp[251][251];
void swap_if_bigger(int& x, int& y)
{
if(x>y)
swap(x, y);
}
//I M N Creates a new table M×N. All the pixels are colored in white (O)
void create()
{
scanf("%d %d", &m, &n);
memset(bmp, 'O', sizeof(bmp));
}
//C Clears the table. The size remains the same. All the pixels became white (O).
void clear()
{
memset(bmp, 'O', sizeof(bmp));
}
//L X Y C Colors the pixel with coordinates (X,Y) in colour C.
void draw_pixel()
{
int x,y;
char color;
scanf("%d %d %c", &x, &y, &color);
bmp[y][x]=color;
}
//V X Y1 Y2 C Draws the vertical segment in the column X between the rows Y1 and Y2 inclusive in colour C.
void draw_ver_seg()
{
int x, y1, y2;
char c;
scanf("%d %d %d %c", &x, &y1, &y2, &c);
swap_if_bigger(y1, y2);
for(int y=y1;y<=y2;y++)
{
bmp[y][x]=c;
}
}
//H X1 X2 Y C Draws the horizontal segment in the row Y between the columns X1 and X2 inclusive in colour C.
void draw_hor_seg()
{
int x1, x2, y;
char c;
scanf("%d %d %d %c", &x1, &x2, &y, &c);
swap_if_bigger(x1, x2);
for(int x=x1;x<=x2;x++)
{
bmp[y][x]=c;
}
}
//K X1 Y1 X2 Y2 C Draws the filled rectangle in colour C. (X1,Y1) is the upper left corner, (X2,Y2) is the lower right corner of the rectangle.
void draw_rect()
{
int x1, y1, x2, y2;
char c;
scanf("%d %d %d %d %c", &x1, &y1, &x2, &y2, &c);
swap_if_bigger(x1, x2);
swap_if_bigger(y1, y2);
for(int x=x1;x<=x2;x++)
for(int y=y1;y<=y2;y++)
{
bmp[y][x]=c;
}
}
//F X Y C Fills 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.
void _fill_region(int x, int y, char c)
{
char old_color=bmp[y][x];
if(old_color==c)
return;
bmp[y][x]=c;
static int direction[4][2]={
{-1, 0},
{1, 0},
{0, -1},
{0, 1},
};
for(int i=0;i<4;i++)
{
int nx=x+direction[i][0];
int ny=y+direction[i][1];
if(nx<1 || nx>m || ny<1 || ny>n)
continue;
if(bmp[ny][nx]==old_color)
{
// cout<<nx<<" "<<ny<<endl;
_fill_region(nx, ny, c);
}
}
}
void fill_region()
{
int x,y;
char c;
scanf("%d %d %c", &x, &y, &c);
_fill_region(x, y, c);
}
//S Name Writes the picture in the file Name.
void save()
{
string filename;
cin>>filename;
cout<<filename<<endl;
for(int y=1;y<=n;y++)
{
for(int x=1;x<=m;x++)
{
cout<<bmp[y][x];
}
cout<<endl;
}
}
int main()
{
char cmd;
char line[256];
while(cin>>cmd)
{
if(cmd=='X')
break;
switch(cmd)
{
case 'I':
create();
break;
case 'C':
clear();
break;
case 'L':
draw_pixel();
break;
case 'V':
draw_ver_seg();
break;
case 'H':
draw_hor_seg();
break;
case 'K':
draw_rect();
break;
case 'F':
fill_region();
break;
case 'S':
save();
break;
default:
gets(line);
continue;
}
}
return 0;
}
PC/UVa 题号: 110105/10267 Graphical Editor (图形化编辑器)题解的更多相关文章
- PC/UVa 题号: 110106/10033 Interpreter (解释器)题解 c语言版
, '\n'); #include<cstdio> #include<iostream> #include<string> #include<algorith ...
- PC/UVa 题号: 110104/706 LC-Display (液晶显示屏)题解
#include <string> #include <iostream> #include <cstring> #include <algorithm> ...
- PC/UVa 题号: 110101/100 The 3n+1 problem (3n+1 问题)
The 3n + 1 problem Background Problems in Computer Science are often classified as belonging to a ...
- 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 MBSubmit: 2 Solved: 2[Submit][Statu ...
- ACM YTU 《挑战编程》第一章 入门 Problem E: Graphical Editor
Description Graphical editors such as Photoshop allow us to alter bit-mapped images in the same way ...
- uva题库爬取
每次进uva都慢的要死,而且一步一步找到自己的那个题目简直要命. 于是,我想到做一个爬取uva题库,记录一下其中遇到的问题. 1.uva题目的链接是一个外部的,想要获取https资源,会报出SNIMi ...
- Asp.Net Core 使用Monaco Editor 实现代码编辑器
在项目中经常有代码在线编辑的需求,比如修改基于Xml的配置文件,编辑Json格式的测试数据等.我们可以使用微软开源的在线代码编辑器Monaco Editor实现这些功能.Monaco Editor是著 ...
- 天大acm 题号1002 Maya Calendar
Description 上周末,M.A. Ya教授对古老的玛雅有了一个重大发现.从一个古老的节绳(玛雅人用于记事的工具)中,教授发现玛雅人使用了一个一年有365天的叫做Haab的历法.这 个Haab历 ...
- The Trip PC/UVa IDs: 110103/10137, Popularity: B, Success rate: average Level: 1
#include<cstdio> #include<iostream> #include<string> #include<algorithm> #in ...
随机推荐
- javascript 命名空间的实例应用
/** * 创建全局对象MYAPP * @module MYAPP * @title MYAPP Global */ var MYAPP = MYAPP || {}; /** * 返回指定的命名空间, ...
- H.264中NAL、Slice与frame意思及相互关系
H.264中NAL.Slice与frame意思及相互关系 NAL nal_unit_type中的1(非IDR图像的编码条带).2(编码条带数据分割块A).3(编码条带数据分割块B).4(编码条带数据分 ...
- 一个P2P点播直播开源项目:P2PCenter
最近跟着公司的项目走,我也研究了不少东西,尤其是在P2P方面,广泛涉猎各种开源项目,尤其是国外的开源项目,意外的发现了一个国内的项目,做的还不错,推荐一下.---------------------使 ...
- JS触发ASP.NET服务器端控件的方法
<asp:Button ID="Button_regId" runat="server" Font-Bold="False" OnCl ...
- C++ STL算法系列3---求和:accumulate
该算法在numeric头文件中定义. 假设vec是一个int型的vector对象,下面的代码: //sum the elements in vec starting the summation wit ...
- 38、FragmentStatePagerAdapter分页
[ ViewPager ] ViewPager 如其名所述,是负责翻页的一个 View.准确说是一个 ViewGroup,包含多个 View 页,在手指横向滑动屏幕时,其负责对 View 进行切换.为 ...
- Delphi RICHEDIT中插入图象
unit InsRich;interfaceuses Windows, Messages, SysUtils, Variants, Classes, Graphics, Controls, Forms ...
- 通过gdb跟踪进程调度分析进程切换的过程
作者:吴乐 山东师范大学 <Linux内核分析>MOOC课程http://mooc.study.163.com/course/USTC-1000029000 本实验目的:通过gdb在lin ...
- Echarts显示全球dns server物理位置
今天YY给了我一大串dns server的ip,然后提出将这些ip物理位置显示在世界地图上(仅仅显示每个地区有几个dns server就好),苦逼了一下午,总算告一段落.把里面关键的点贴上来以后看.. ...
- 基于Maven管理的Mapreduce程序下载依赖包到LIB目录
1.Mapreduce程序需要打包作为作业提交到Hadoop集群环境运行,但是程序中有相关的依赖包,如果没有一起打包,会出现xxxxClass Not Found . 2.在pom.xml文件< ...