ZOJ-1067 基本运算题

1067：给出16种RGB值作为基准，将后来的颜色值映射到距离最近的颜色。
距离的定义为



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)



简单题。依次计算找距离最小即可。

#include<stdio.h>
#include<iostream>
using namespace std;

int color[16][3];
int main()
{
	int R,G,B;
	int index;
	int diff;
	int cal;

	//目标颜色
	for(int i=0;i<16;i++)
	{
		cin>>color[i][0];
		cin>>color[i][1];
		cin>>color[i][2];
	}

	while(1)
	{
		index=-1;
		diff=-1;
		cin>>R;
		cin>>G;
		cin>>B;
		
		if(R==-1&&G==-1&&B==-1)
			break;

		for(int i=0;i<16;i++)
		{
			cal=(R-color[i][0])*(R-color[i][0])+(G-color[i][1])*(G-color[i][1])+(B-color[i][2])*(B-color[i][2]);
			if(diff==-1||cal<diff)
			{
				diff=cal;
				index=i;
			}
		}

		printf("(%d,%d,%d) maps to (%d,%d,%d)\n",R,G,B,color[index][0],color[index][1],color[index][2]);

	}


	
}