1067_Color Me Less
来源:互联网 发布:复制手机联系人软件 编辑:程序博客网 时间:2024/05/17 04:15
Problem
A color reduction is a mapping from a set of discrete colors to a smaller one. The solution to this problem requires that you perform just such a mapping in a standard twenty-four bit RGB color space. The input consists of a target set of sixteen RGB color values, and a collection of arbitrary RGB colors to be mapped to their closest color in the target set. For our purposes, an RGB color is defined as an ordered triple (R,G,B) where each value of the triple is an integer from 0 to 255. The distance between two colors is defined as the Euclidean distance between two three-dimensional points. That is, given two colors (R1,G1,B1) and (R2,G2,B2), their distance D is given by the equation
The input file is a list of RGB colors, one color per line, specified as three integers from 0 to 255 delimited by a single space. The first sixteen colors form the target set of colors to which the remaining colors will be mapped. The input is terminated by a line containing three -1 values.
Output
For each color to be mapped, output the color and its nearest color from the target set.
Example
Input
0 0 0
255 255 255
0 0 1
1 1 1
128 0 0
0 128 0
128 128 0
0 0 128
126 168 9
35 86 34
133 41 193
128 0 128
0 128 128
128 128 128
255 0 0
0 1 0
0 0 0
255 255 255
253 254 255
77 79 134
81 218 0
-1 -1 -1
Output
(0,0,0) maps to (0,0,0)
(255,255,255) maps to (255,255,255)
(253,254,255) maps to (255,255,255)
(77,79,134) maps to (128,128,128)
(81,218,0) maps to (126,168,9)
*****************************************************************************************************************************************************************************************
#include<iostream>
#include<cmath>using namespace std;
int main()
{ const int nn=48;
int target[nn];
int color[3];
int flag;
double resault=500;
double a1,a2,a3,b1,b2,b3,cc;
for(int i=0;i!=nn;++i)
cin>>target[i];
while(1)
{
for(int i=0;i!=3;i++)
cin>>color[i];
if(color[0]==-1&&color[1]==-1&&color[2]==-1)
break;
resault=500;//每比较一组置一次值
for(int i=0;i!=16;i++)
{
a1=color[0]-target[i*3];
a2=color[1]-target[i*3+1];
a3=color[2]-target[i*3+2];
b1=pow(a1,2);
b2=pow(a2,2);
b3=pow(a3,2);
cc=sqrt(b1+b2+b3);
if(cc<resault)
{
resault=cc;
flag=i;
}
}
cout<<"("<<color[0]<<","<<color[1]<<","<<color[2]<<") maps to ("<<target[flag*3]<<","<<target[flag*3+1]<<","<<target[flag*3+2]<<")"<<endl;
}
return 0;
}
- 1067_Color Me Less
- 1067 Color Me Less
- ZOJ-1067-Color Me Less
- zoj 1067 Color Me Less
- ZOJ 1067Color Me Less
- zoj 1067 Color Me Less
- ZOJ 1067 Color Me Less
- zoj 1067 Color Me Less
- ZOJ 1067 Color Me Less
- ZOJ 1067 :Color Me Less
- zoj--1067--Color Me Less
- zoj 1067 Color Me Less
- zoj 1067 Color Me Less
- zoj 1067 Color Me Less
- zoj 1067 Color Me Less
- ZOJ--1067:Color Me Less
- Solution of ZOJ 1067 Color Me Less
- php - zoj 1067 Color Me Less
- 博客正式版家
- 1151_Word Reversal
- 2012年年终总结(非官方版)
- java实现Hbase中的查询(一)Filter方式
- iOS中使用RegexKitLite来试用正则表达式
- 1067_Color Me Less
- 参考:MyEclipse10+Flex4.6+BlazeDS终于可以跑程序了
- java实现hbase表创建、数据插入、删除表
- Spring 3.1MVC 实战
- 鉴客 Android 背景图片重复 Background repeat
- Reflector 已经out了,试试ILSpy
- Air Mobile As3 App
- 国际著名黑客大赛介绍与比较
- 1051_A New Growth Industry