zoukankan      html  css  js  c++  java
  • Color Me Less

    Color Me Less

    Time Limit: 2 Seconds      Memory Limit: 65536 KB

    Problem

    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

     

     1 #include <iostream>
     2 #include <string>
     3 #include <algorithm>
     4 #include <map>
     5 #include <vector>
     6 #include <cstdio>
     7 #include <cmath>
     8 #include <cstring>
     9 using namespace std;
    10 
    11 int main()
    12 {
    13     int r[16], b[16], g[16];
    14     for(int i = 0; i < 16; i++)
    15     {
    16         scanf("%d %d %d", r + i, g + i, b + i);
    17     }
    18     int x, y, z;
    19     while(scanf("%d %d %d", &x, &y, &z))
    20     {
    21         if(x == -1 && y == -1 && z == -1) break;
    22         double minD = (x - r[0]) * (x - r[0]) + (y - g[0])* (y - g[0]) + (z - b[0]) * (z - b[0]);
    23         int id = 0;
    24         for(int i = 1; i < 16; i++)
    25         {
    26             double D = (x - r[i]) * (x - r[i]) + (y - g[i]) * (y - g[i]) + (z - b[i]) * (z - b[i]);
    27             if(minD > D)
    28             {
    29                 minD = D;
    30                 id = i;
    31             }
    32         }
    33         printf("(%d,%d,%d) maps to (%d,%d,%d)
    ", x, y, z, r[id], g[id], b[id]);
    34       //  cout << id << endl;
    35     }
    36     return 0;
    37 }
  • 相关阅读:
    CrawlSpiders
    从抓取Tencent中学习Scrapy
    对象返回规范的url的两种方式的两种方式
    多对多关系的额外字段
    Django定时任务
    Scripy学习(一)
    Django开发博客一(搭建模型和准备数据)
    求并集
    求子集、交集
    java数学函数Math类中常用的方法
  • 原文地址:https://www.cnblogs.com/cszlg/p/3246745.html
Copyright © 2011-2022 走看看