POJ1046 ZOJ1076 UVALive2281 Color Me Less【暴力】

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于 2018-07-01 10:15:55 发布

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分类专栏： # ICPC-暴力枚举 # ICPC-POJ # ICPC-UVALive # ICPC-ZOJ 文章标签： POJ1046 ZOJ1076 UVALive2281 Color Me Less Color Me Less Color Me Less Color Me Less

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本文介绍了一个颜色映射问题，该问题需要将任意RGB颜色映射到一个预定义的颜色集中最接近的颜色。通过计算不同颜色之间的欧几里得距离并选择最小距离对应的预定义颜色来实现映射。

Color Me Less

Time Limit: 1000MS		Memory Limit: 10000K
Total Submissions: 35455		Accepted: 17187

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)

Source

Greater New York 2001

Regionals 2001 >> North America - Greater NY

问题链接：POJ1046 ZOJ1076 UVALive2281 Color Me Less

问题描述：（略）

问题分析：

这个问题用枚举来解决，计算所有的距离再求个最值就可以了。开平方计算是不需要的，可以加快计算速度，但是需要注意最小值的初值设置。

程序说明：（略）

参考链接：（略）

题记：（略）

AC的C++语言程序如下：

/* POJ1046 ZOJ1076 UVALive2281 Color Me Less */

#include <iostream>
#include <stdio.h>

using namespace std;

const int N = 16;
struct Color {
    int r, g, b;
} c[N];

int main()
{
    for(int i = 0; i < N; i++)
        scanf("%d%d%d", &c[i].r, &c[i].g, &c[i].b);

    int r, g, b;
    while(~scanf("%d%d%d", &r, &g, &b) && (r != -1 || g != -1 || b != -1)) {
        int minc = 255 * 255 * 3, k = 0;
        for(int i = 0; i < N; i++) {
            int t = (c[i].r - r) * (c[i].r - r) + (c[i].g - g) * (c[i].g - g) + (c[i].b - b) * (c[i].b - b);
            if(t < minc)
                minc = t, k = i;
        }

        printf("(%d,%d,%d) maps to (%d,%d,%d)\n", r, g, b, c[k].r, c[k].g, c[k].b);
    }

    return 0;
}