Color me less

最新推荐文章于 2023-02-21 14:17:57 发布

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本文介绍了一个基于欧几里得距离的颜色映射问题解决方案。该方案使用C语言实现，旨在将RGB颜色空间中的任意颜色映射到一个预定义的16种目标颜色之一，通过计算每种目标颜色与待映射颜色之间的距离来确定最接近的目标颜色。

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Color Me Less
Time Limit: 1000MS
Memory Limit: 10000K
Total Submissions: 26432
Accepted: 12740
Description

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

Input

The input 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.

If there are more than one color with the same smallest distance, please output the color given first in the color set.
Sample 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
Sample 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<stdio.h>
int main()
    {
    int save[17][4];
    int clo[4];
    int n,i;
    for(i=1;i<=16;i++)
    {
        for(n=1;n<=3;n++)
        scanf("%d",&save[i][n]);
    }

    while(1)
    {
        int ne[3];
        scanf("%d %d %d",&ne[1],&ne[2],&ne[3]);
        if(ne[1]==-1&&ne[2]==-1&&ne[3]==-1)
        break;
        int min=3*255*255;
        for(i=1;i<=16;i++)
        {
            int d;
            d=(save[i][1]-ne[1])*(save[i][1]-ne[1])+(save[i][2]-ne[2])*(save[i][2]-ne[2])+(save[i][3]-ne[3])*(save[i][3]-ne[3]);
            if(d<min)
            {
            min=d;
            clo[1]=save[i][1];
            clo[2]=save[i][2];
            clo[3]=save[i][3];
            }
        }
        printf("(%d,%d,%d) maps to (%d,%d,%d)\n",ne[1],ne[2],ne[3],clo[1],clo[2],clo[3]);

    }

    return 0;

}