Color Me Less

题目连接

题目介绍：

D - Color Me Less

Time Limit:1000MS     Memory Limit:10000KB     64bit IO Format:%I64d & %I64u

Submit Status

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)

AC代码
#include <stdio.h>
#include <stdlib.h>
//#include <math.h>

int  a[1000],b[1000],c[1000];
int x,y,z,n,i;
int main ()
{
	float sum,count;
	
	for(i=0;i<16;i++)
	{
		scanf("%d %d %d",&a[i],&b[i],&c[i]);
	}
	while(scanf("%d %d %d",&x,&y,&z),(x!=-1 && y!=-1 && z!=-1))//都为-1时退出
	{
		sum =0;
		count = (1<<30);//定义count 与sum比较  把最短距离挑选出来 其中count不可以过小不可以小于255的三次方
		for(i=0;i<16;i++)
		{
			sum=((a[i]-x)*(a[i]-x)+(b[i]-y)*(b[i]-y)+(c[i]-z)*(c[i]-z));//既然sqrt可以比较大小自然它的平方也可以比较
			if(count >sum)
			{
				count =sum;
				n=i;
			}
		}
		printf("(%d,%d,%d) maps to (%d,%d,%d)
",x,y,z,a[n],b[n],c[n]);//输出
	}
	return 0;
}