标签:des style http color strong os
Time Limit: 1000MS | Memory Limit: 10000K | |
Total Submissions: 30146 | Accepted: 14634 |
Description
Input
Output
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)
Source
解题思路:
一种颜色用三元组(r,g,b)表示,两个颜色的距离为。给出16个已知的颜色值,求这里面距离给定的颜色值距离最短的颜色值。
枚举,比较距离就可以了。
代码:
#include <iostream> #include <cmath> const int inf=0x7fffffff; using namespace std; struct RGB { double r,g,b; int match;//用来记录距离最短的颜色是第几个 }rgb[100]; int main() { int k=1; while(cin>>rgb[k].r>>rgb[k].g>>rgb[k].b&&rgb[k].r!=-1&&rgb[k].g!=-1&&rgb[k].b!=-1) { k++; } double dis; for(int i=17;i<=k-1;i++) { dis=inf; for(int j=16;j>=1;j--) { double temp=sqrt((rgb[i].r-rgb[j].r)*(rgb[i].r-rgb[j].r)+(rgb[i].g-rgb[j].g)*(rgb[i].g-rgb[j].g)+(rgb[i].b-rgb[j].b)*(rgb[i].b-rgb[j].b)); if(dis>=temp) { dis=temp;//最短距离 rgb[i].match=j;//第i中颜色距离第j种颜色距离最短 } } } for(int i=17;i<=k-1;i++) { cout<<"("<<rgb[i].r<<","<<rgb[i].g<<","<<rgb[i].b<<") maps to ("<<rgb[rgb[i].match].r<<","<<rgb[rgb[i].match].g<<","<<rgb[rgb[i].match].b<<")"<<endl; } return 0; }
[ACM] POJ 1046 Color Me Less,布布扣,bubuko.com
标签:des style http color strong os
原文地址:http://blog.csdn.net/sr_19930829/article/details/37727653