Color Me Less

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 0255 255 2550 0 11 1 1128 0 00 128 0128 128 00 0 128126 168 935 86 34133 41 193128 0 1280 128 128128 128 128255 0 00 1 00 0 0255 255 255253 254 25577 79 13481 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)
题意：从第十六行开始计算这个点与前十六个点的距离，找出最小的。枚举一遍就好了。
#include<iostream>#include<cstdio>#include<cstring>using namespace std;int main(){    int a[10000];    int b[10000];    int c[10000];    int i=0,t,k,j;    while(scanf("%d %d %d",&a[i],&b[i],&c[i])!=EOF)    {        int min1=1000000000;        if(a[i]==-1&&b[i]==-1&&c[i]==-1)            break;        if(i>=16)        {            for(j=0; j<16; j++)            {                t=((a[i]-a[j])*(a[i]-a[j])+(b[i]-b[j])*(b[i]-b[j])+(c[i]-c[j])*(c[i]-c[j]));                if(t<min1)                {                    k=j;                    min1=t;                }            }            printf("(%d,%d,%d) maps to (%d,%d,%d)\n",a[i],b[i],c[i],a[k],b[k],c[k]);        }        i++;    }}