2017 ACM-ICPC 亚洲区（南宁赛区）网络赛-I-GSM Base Station Identification(线性变换)

In the Personal Communication Service systems such as GSM (Global System for Mobile Communications), there are typically a number of base stations spreading around the service area. The base stations are arranged in a cellular structure, as shown in the following figure. In each cell, the base station is located at the center of the cell.

For convenience, each cell is denoted by [$i$, $j$]. The cell covers the origin is denoted by [$0$, $0$]. The cell in the east of [$0$, $0$] is denoted by [$1$, $0$]. The cell in the west of [$0$, $0$] is denoted by [$-1$, $0$]. The cell in the northeast of [$0$, $0$] is denoted by [$0$, $1$]. The cell in the southwest of [$0$, $0$] is denoted by [$0$, $-1$]. This notation can be easily generalized, as shown in the above figure.

Now the question is as follows. We have a service area represented by a Euclidean plane (i.e., $x-y$ plane). Each unit is $1$ Km. For example, point ($5$, $0$) in the plane means the location at a distance of $5$ Km to the east of the origin. We assume that there are totally $400$ cells, denoted by [$i$, $j$], $i=-9...10$, $j=-9...10$. The base station of cell [$0$, $0$] is located at the origin of the Euclidean plane. Each cell has a radius of $R$ = $5$ Km, as shown in the following figure.

You are given an input ($x$, $y$), which indicates a mobile phone’s location. And you need to determine the cell [$i$, $j$] that covers this mobile phone and can serve this phone call.

For example, given a location ($10$, $0$), your program needs to output the cell [$1$, $0$], which can cover this location. Specifically, the input and output are:

• input = ($x$, $y$). hhis is a location on the Euclidean plane. This value will not exceed the service area covered by the $400$ cells. That is, you do not need to handle the exceptional case that the input is out of the boundary of the service area.
• output = [$i$, $j$]. One of the $400$ cells that covers location [$i$, $j$]

Input Format

A list of $10$ locations.

Output Format

A list of $10$ cells covering the above $10$ locations in the correct order.

Please be reminded that there exist a space between coordinates.

样例输入

1 00 152 013 75 510 1525 15-13 -812 -7-10 0

样例输出

[0,0], [-1,2], [0,0], [1,1], [0,1], [0,2], [2,2], [-1,-1], [2,-1], [-1,0]

题目来源

2017 ACM-ICPC 亚洲区（南宁赛区）网络赛

题意：给你10个坐标，问你这10个点在哪个正六边形内。

题解：一开始感觉是个超级麻烦的暴力啊，比赛时室友做的这道题，说是线性变换，然而线代早忘得

一干二净了，用线性变换的话就很简单了，将每个正六边形的中心点变换到普通的二维坐标系上就OK了。

我们可以看图找出变换关系式，剩下的就是再正常不过的枚举了。。。。

#include<set>  #include<map>     #include<stack>            #include<queue>            #include<vector>    #include<string> #include<time.h>#include<math.h>            #include<stdio.h>            #include<iostream>            #include<string.h>            #include<stdlib.h>    #include<algorithm>   #include<functional>    using namespace std;            #define ll long long       #define inf 1000000000     #define mod 1000000007             #define maxn  1360100#define lowbit(x) (x&-x)            #define eps 1e-9int a[15],b[15];int main(void){int i,j,t;double x,y,xx,yy;for(t=1;t<=10;t++){scanf("%lf%lf",&x,&y);x/=5;y/=5;for(i=-9;i<=10;i++)for(j=-9;j<=10;j++){yy=3.0/2*j;xx=i*sqrt(3.0)+j*sqrt(3.0)/2;if((xx-x)*(xx-x)+(yy-y)*(yy-y)<=1)a[t]=i,b[t]=j;}}for(i=1;i<10;i++)printf("[%d,%d], ",a[i],b[i]);printf("[%d,%d]\n",a[10],b[10]);return 0;}