[ACM] POJ 1046 Color Me Less

Color Me Less

Time Limit: 1000MS Memory Limit: 10000KTotal Submissions: 30146 Accepted: 14634

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 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)

Source

Greater New York 2001

 

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，比较距离就可以了。

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;}