Color Me Less (poj 1046) (sdut 1047)水题

Color Me Less

Time Limit: 1000ms Memory limit: 32765K 有疑问？点这里^_^

题目描述

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.

输入

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.

输出

For each color to be mapped, output the color and its nearest color from the target set.

示例输入

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

示例输出

(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)

题意：先输入16个点，然后再输入若干个点，在前边的16个点中找到与其距离最小的点

#include <iostream>
#include <cstdio>
#include <cstring>
#include <cstdlib>
#include <cmath>
using namespace std;
struct node
{
int r,g,b;
}col[16],q;
int main()
{
int i,j,k;
double d,D;
int r1,g1,b1;
for(i=0;i<16;i++)
cin>>col[i].r>>col[i].g>>col[i].b;
while(cin>>q.r>>q.g>>q.b)
{
if(q.r==-1&&q.g==-1&&q.b==-1)
break;//输入为三个-1时结束
else
{
d=500;
for(i=0;i<16;i++)
{
r1= (q.r-col[i].r)*(q.r-col[i].r);
g1=(q.g-col[i].g)*(q.g-col[i].g);
b1=(q.b-col[i].b)*(q.b-col[i].b);
D=sqrt(r1+g1+b1);//求两点间的距离
if(D<d)
{
d=D;
j=i;
}
}
cout<<"("<<q.r<<","<<q.g<<","<<q.b<<")" <<" maps to ";
cout<<"("<<col[j].r<<","<<col[j].g<<","<<col[j].b<<")" <<endl;
}
}
return 0;
}