Color Me Less，哪里出问题了？
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 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.
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)
我写的代码：
#include<stdio.h>
#include<math.h>
const int MAXSIZE=100;
int pos[100],i,j;
struct RGB
{
double red;
double green;
double blue;
}color[MAXSIZE];
double distance(double color[i],double color[j])
{
double r=color[i].red-color[j].red;
double g=color[i].green-color[j].green;
double b=color[i].blue-color[j].blue;
return sqrt(r*r+g*g+b*b);
}
int main()
{
double r,g,b,dist;
int count=0,k;
while(scanf("%lf %lf %lf",&r,&g,&b)!=EOF && r!=-1 && g!=-1 && b!=-1)
{
color[count].red=r;
color[count].green=g;
color[count].blue=b;
++count;
}
double min;
k=0;
for(i=16;i<count;i++)
{
min=99999;
for(j=0;j<16;j++)
{
dist=distance(color[i],color[j]);
if(dist<min)
{
min=dist;
pos[k]=j;
}
}
k++;
}
for(i=16,k=0;i<count;i++,k++)
{
printf("(%lf,%lf,%lf) maps to ",color[i].red,color[i].green,color[i].blue);
printf("(%lf,%lf,%lf)\n",color[pos[k]].red,color[pos[k]].green,color[pos[k]].blue);
}
return 0;
}
但是编译错误，好像是函数调用出错了，请问到底哪里出问题了？
[解决办法]
这个题目的意思是：
求在前16个点中找出与目标点最近的那个点，如果有多个相同点则输出第一个
double distance(RGB aa,RGB bb) //函数形参搞错{ double r=aa.red-bb.red; double g=aa.green-bb.green; double b=aa.blue-bb.blue; return sqrt(r*r+g*g+b*b);}