Color Me Less（水题）

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 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

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)

题意：这相当于是在一个三维坐标里求点之间的最小距离。先输入十六个点，从第17行到文件结束   找每一行的点距离上面十六行距离最短的点   然后输出。

#include <stdio.h>
#include <string.h>
#include <stdlib.h>
#include <iostream>
#include <algorithm>
using namespace std;
#define inf 999999999
struct node
{
    int r,g,b;
}cor1[16],cor2;
int main()
{
    int i;
    int dis,mdis;
    int flag;
    for(i=0;i<16;i++)
    {
        scanf("%d %d %d",&cor1[i].r,&cor1[i].g,&cor1[i].b);
    }
    while(1)
    {
        scanf("%d %d %d",&cor2.r,&cor2.g,&cor2.b);
        mdis=inf;
        if(cor2.r==-1&&cor2.g==-1&&cor2.b==-1)
            break;
    for(i=0;i<16;i++)
    {
        dis=(cor1[i].r-cor2.r)*(cor1[i].r-cor2.r)+(cor1[i].g-cor2.g)*(cor1[i].g-cor2.g)+(cor1[i].b-cor2.b)*(cor1[i].b-cor2.b);
        if(mdis>dis)
        {
            mdis=dis;
            flag=i;
        }
            }
    printf("(%d,%d,%d) maps to (%d,%d,%d)\n",cor2.r,cor2.g,cor2.b,cor1[flag].r,cor1[flag].g,cor1[flag].b);
    }
    return 0;
}