POJ 1046(颜色映射 简单数学) 解题报告
/*____________________________________________POJ 1046题_____________________________________________________ Color Me Less Time Limit: 1000MS Memory Limit: 10000K Total Submissions: 21650 Accepted: 10300 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 D=sqrt((R2-R1)*(R2-R1)+(G2-G1)*(G2-G1)+(B2-B1)*(B2-B1)) 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) ____________________________________________________________________________________________________________*/ #include<stdio.h> #include<math.h> int main() { int i,j,t1,t2,t3; float d,sd; //sd是最小距离 int target[16][3], color[3], map[3]; // FILE *fin=fopen("input.txt","r"); for(i=0;i<16;i++) for(j=0;j<3;j++) // fscanf(fin,"%d",&target[i][j]); scanf("%d",&target[i][j]); while(1) { for(i=0;i<3;i++) scanf("%d",&color[i]); // fscanf(fin,"%d",&color[i]); if(color[0]==-1 && color[1]==-1 && color[2]==-1) break; t1=target[0][0]-color[0]; t2=target[0][1]-color[1]; t3=target[0][2]-color[2]; sd=sqrt((float)(t1*t1 + t2*t2 + t3*t3)); map[0]=target[0][0]; map[1]=target[0][1]; map[2]=target[0][2]; for(i=1;i<16;i++) { t1=target[i][0]-color[0]; t2=target[i][1]-color[1]; t3=target[i][2]-color[2]; d=sqrt((float)(t1*t1 + t2*t2 + t3*t3)); if(d<sd) { sd=d; map[0]=target[i][0]; map[1]=target[i][1]; map[2]=target[i][2]; } } printf("(%d,%d,%d) maps to (%d,%d,%d)/n",color[0],color[1],color[2],map[0],map[1],map[2]); } return 0; } //发现gcc编译器中若用整形做sqrt()的参数，编译器无法区分重载(double float)，所以还要先进行类型转换