zoukankan      html  css  js  c++  java
  • ZOJ Problem Set

    ZOJ Problem Set - 1067
    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

    The input file 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.


    Example

    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

    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)

    AC 代码:

    #include<iostream>
    #include<stdio.h>
    #include<cmath>
    using namespace std;
    int t[16][3];
    void table()
    {
     for(int i=0;i<16;i++)
     {
      for(int j=0;j<3;j++)
      {
       cin>>t[i][j];
      }
     }
    }
    int main()
    {
        table();
     int r,g,b;
     while(cin>>r>>g>>b)
     {
        if(r==-1||g==-1||b==-1)
           break;
        double d=0.0;
        double min=10000000;
        int tr,tg,tb;
        for(int i=0;i<16;i++)
        {
         d=(r-t[i][0])*(r-t[i][0])+(g-t[i][1])*(g-t[i][1])+(b-t[i][2])*(b-t[i][2]);
         if(min>d)
         {
          min=d;
          tr=t[i][0];
          tg=t[i][1];
          tb=t[i][2];
         }
        }
        printf("(%d,%d,%d) maps to (%d,%d,%d) ",r,g,b,tr,tg,tb);
     } 
    }

  • 相关阅读:
    poj 1061 青蛙的约会 ——扩展欧几里得
    2013年4月4日 雨
    hdu3555 Bomb ——数位DP入门题
    2013年4月6日四校联赛总结
    speedcell's SPFA
    nefu118 n!后面有多少个0 数论
    zoj 3409 KKV
    uva 11991 Easy Problem from Rujia Liu?
    zoj 1649 Rescue ——BFS入门题
    poj 3233 Matrix Power Series ——矩阵快速幂+二分求解
  • 原文地址:https://www.cnblogs.com/jackwuyongxing/p/3366513.html
Copyright © 2011-2022 走看看