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2013
11-09

POJ 1046 Color Me Less [解题报告] Java

Color Me Less

问题描述 :

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

输入:

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.

输出:

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.

样例输入:

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

样例输出:

(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)

解题代码:

import java.util.*;   
  
public class Main {   
  
    public static void main(String[] args) {   
        Scanner cin = new Scanner(System.in);   
        String[] str;   
           
        int[][] target = new int[16][3];   
        int[] temp = new int[3];   
        int minIndex = 0;   
        double minValue = -1;   
        double tempValue = 0;   
           
        for(int i = 0; i < 16; i++)   
        {   
            str = cin.nextLine().split(" ");   
            target[i][0] = Integer.valueOf(str[0]).intValue();   
            target[i][1] = Integer.valueOf(str[1]).intValue();   
            target[i][2] = Integer.valueOf(str[2]).intValue();   
        }   
           
        while(cin.hasNext())   
        {   
            minIndex = 0;   
            minValue = -1;   
            tempValue = 0;   
            str = cin.nextLine().split(" ");   
            temp[0] = Integer.valueOf(str[0]).intValue();   
            temp[1] = Integer.valueOf(str[1]).intValue();   
            temp[2] = Integer.valueOf(str[2]).intValue();   
               
            if(temp[0]==-1 && temp[1]==-1 && temp[2]==-1)   
                break;   
               
            for(int i = 0; i < 16; i++)   
            {   
                tempValue = getEuclideanD(temp, target[i]);   
                if(tempValue == 0)   
                {   
                    minValue = tempValue;   
                    minIndex = i;   
                    break;   
                }   
                if(minValue == -1)   
                {   
                    minValue = tempValue;   
                    minIndex = 0;   
                    continue;   
                }   
                if(minValue > tempValue)   
                {   
                    minValue = tempValue;   
                    minIndex = i;   
                }   
            }   
               
            System.out.println(   
                    "("+temp[0]+","+temp[1]+","+temp[2]+   
                    ") maps to "+   
                    "("+target[minIndex][0]+","+   
                    target[minIndex][1]+","+   
                    target[minIndex][2]+")");   
        }   
    }   
       
    private static double getEuclideanD(int[] p1, int[] p2)   
    {   
        double value = 0;   
           
        value = Math.sqrt(Math.pow((p2[0]-p1[0]), 2) +   
                    Math.pow((p2[1]-p1[1]), 2) +    
                    Math.pow((p2[2]-p1[2]), 2));   
           
        return value;   
           
    }   
  
}

  1. 一开始就规定不相邻节点颜色相同,可能得不到最优解。我想个类似的算法,也不确定是否总能得到最优解:先着一个点,随机挑一个相邻点,着第二色,继续随机选一个点,但必须至少有一个边和已着点相邻,着上不同色,当然尽量不增加新色,直到完成。我还找不到反例验证他的错误。。希望LZ也帮想想, 有想法欢迎来邮件。谢谢

  2. 在方法1里面:

    //遍历所有的边,计算入度
    for(int i=0; i<V; i++)
    {
    degree = 0;
    for (j = adj .begin(); j != adj .end(); ++j)
    {
    degree[*j]++;
    }
    }

    为什么每遍历一条链表,要首先将每个链表头的顶点的入度置为0呢?
    比如顶点5,若在顶点1、2、3、4的链表中出现过顶点5,那么要增加顶点5的入度,但是在遍历顶点5的链表时,又将顶点5的入度置为0了,那之前的从顶点1234到顶点5的边不是都没了吗?