zju1067题解
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题目:
Color Me Less
Time Limit: 1 Second Memory Limit: 32768 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)
Source: Greater New York 2001
题目大意:输入前16行为目标颜色组,是一个有序三元组,之后输入多行颜色组,如果是三个-1,表示结束,目标颜色组的三个颜色设为r1,g1,b1,其余的颜色设为r2,g2,b2,两者之间的距离公式为d=sqrt(pow(r2-r1,2)+pow(g2-g1,2)+pow(b2-b1,2)),输出与后输入颜色组距离最接近的目标颜色组中的颜色,。
算法分析:其实本题就是求后输入的颜色与目标颜色组中d最小的颜色,照着题目中的公式计算就可以了。我的方法就是先求出16个目标颜色组中的颜色与输入颜色的d,在求出最小的d,最后找出最小的d对应的目标颜色。
代码:
语言:c
#include<stdio.h>
#include<math.h> int main() { int a[16][3],colour[3]; float min,distant[16]; int i,j; for(i=0;i<16;++i) for(j=0;j<3;++j) scanf("%d",&a[i][j]); while(1) { for(i=0;i<3;++i) scanf("%d",&colour[i]); if(colour[0]==-1&&colour[1]==-1&&colour[2]==-1) break; for(i=0;i<16;++i) distant[i]=sqrt(pow((colour[0]-a[i][0]),2)+pow((colour[1]-a[i][1]),2)+pow((colour[2]-a[i][2]),2)); min=distant[0]; for(i=0;i<16;++i) if(distant[i]<=min) min=distant[i]; for(i=0;i<16;++i) { if(min==distant[i]) { printf("(%d,%d,%d) maps to (%d,%d,%d)/n",colour[0],colour[1],colour[2],a[i][0],a[i][1],a[i][2]); break; } } } return 0; }
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