POJ 1046-Color Me Less【暴力】

Bertha 。 2022-08-20 14:28 17阅读 0赞

C - Color Me Less

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

Submit [ Status][Status] [ Practice][Practice] [ POJ 1046][POJ 1046]

Appoint description: A1Dark (2013-10-07) System Crawler (2016-02-26)

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  
![6a8eebe9d04c25e300a89793f69398d7][]

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)

解题思路：

输入16个坐标，再输入的要找到离他最近的点。

#include<stdio.h>
    #include<string.h>
    #include<algorithm>
    #define INF 0x3f3f3f3f
    using namespace std;
    struct node
    {
    	int x,y,z;
    }wc[100000];
    void ji(int xx,int yy,int zz)
    {
    	int zhi=INF,wei=0;
    	int i,j;
    	for(i=0;i<16;i++)
    	{
    		int wc1=((wc[i].x-xx)*(wc[i].x-xx)+(wc[i].y-yy)*(wc[i].y-yy)+(wc[i].z-zz)*(wc[i].z-zz));
    		if(wc1<zhi)
    		{
    			wei=i;
    			zhi=wc1;
    		}
    	}
    	printf("(%d,%d,%d)\n",wc[wei].x,wc[wei].y,wc[wei].z);
    }
    int main()
    {
    	int i,j;
    	for(i=0;i<16;i++)
    	{
    		scanf("%d%d%d",&wc[i].x,&wc[i].y,&wc[i].z);
    	}
    	while(1)
    	{
    		int xx,yy,zz;
    		scanf("%d%d%d",&xx,&yy,&zz);
    		if(xx==-1&&yy==-1&&zz==-1)
    		{
    			break;
    		}
    		printf("(%d,%d,%d) maps to ",xx,yy,zz);
    		ji(xx,yy,zz);
    	}
    	return 0;
    }

[Status]: http://acm.hust.edu.cn/vjudge/contest/view.action?cid=108073#status//C/0
[Practice]: http://acm.hust.edu.cn/vjudge/problem/viewProblem.action?id=16362
[POJ 1046]: http://acm.hust.edu.cn/vjudge/problem/visitOriginUrl.action?id=16362
[6a8eebe9d04c25e300a89793f69398d7]: http://7xjob4.com1.z0.glb.clouddn.com/6a8eebe9d04c25e300a89793f69398d7