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

Output

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)

Source

解题思路：

一种颜色用三元组（r,g,b)表示。两个颜色的距离为。

给出16个已知的颜色值，求这里面距离给定的颜色值距离最短的颜色值。

枚举，比較距离就能够了。

代码：

#include <iostream>

#include <cmath>

const int inf=0x7fffffff;

using namespace std;

struct RGB

{

    double r,g,b;

    int match;//用来记录距离最短的颜色是第几个

}rgb[100];

int main()

{

    int k=1;

    while(cin>>rgb[k].r>>rgb[k].g>>rgb[k].b&&rgb[k].r!=-1&&rgb[k].g!=-1&&rgb[k].b!=-1)

    {

        k++;

    }

    double dis;

    for(int i=17;i<=k-1;i++)

    {

        dis=inf;

        for(int j=16;j>=1;j--)

        {

            double temp=sqrt((rgb[i].r-rgb[j].r)*(rgb[i].r-rgb[j].r)+(rgb[i].g-rgb[j].g)*(rgb[i].g-rgb[j].g)+(rgb[i].b-rgb[j].b)*(rgb[i].b-rgb[j].b));

            if(dis>=temp)

                {

                    dis=temp;//最短距离

                    rgb[i].match=j;//第i中颜色距离第j种颜色距离最短

                }

        }

    }

    for(int i=17;i<=k-1;i++)

    {

        cout<<"("<<rgb[i].r<<","<<rgb[i].g<<","<<rgb[i].b<<") maps to ("<<rgb[rgb[i].match].r<<","<<rgb[rgb[i].match].g<<","<<rgb[rgb[i].match].b<<")"<<endl;

    }

    return 0;

}