2017 ACM-ICPC 亚洲区（南宁赛区）网络赛 GSM Base Station Identification (点在多边形内模板)...

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概述

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 [

Now the question is as follows. We have a service area represented by a Euclidean plane (i.e.,

You are given an input (

For example, given a location (

input = (
output = [

Input Format

A list of

Output Format

A list of

Please be reminded that there exist a space between coordinates.

样例输入

样例输出

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

  1 #include <bits/stdc++.h>
  2 using namespace std;
  3 const double inf = 1000000000.0;
  4 const double ESP = 1e-5;
  5 const int MAX_N = 1000;
  6 const double o = 2.5*sqrt(3.0);
  7 struct Point
  8 {
  9     double x, y;
 10 };
 11 struct LineSegment
 12 {
 13     Point pt1, pt2;
 14 };
 15 typedef vector<Point> Polygon;
 16 Polygon pp;
 17 double Multiply(Point p1, Point p2, Point p0)
 18 {
 19     return ( (p1.x - p0.x) * (p2.y - p0.y) - (p2.x - p0.x) * (p1.y - p0.y) );
 20 }
 21 bool IsOnline(Point point, LineSegment line)
 22 {
 23     return( ( fabs(Multiply(line.pt1, line.pt2, point)) < ESP ) &&
 24             ( ( point.x - line.pt1.x ) * ( point.x - line.pt2.x ) <= 0 ) &&
 25             ( ( point.y - line.pt1.y ) * ( point.y - line.pt2.y ) <= 0 ) );
 26 }
 27 bool Intersect(LineSegment L1, LineSegment L2)
 28 {
 29     return( (max(L1.pt1.x, L1.pt2.x) >= min(L2.pt1.x, L2.pt2.x)) &&
 30             (max(L2.pt1.x, L2.pt2.x) >= min(L1.pt1.x, L1.pt2.x)) &&
 31             (max(L1.pt1.y, L1.pt2.y) >= min(L2.pt1.y, L2.pt2.y)) &&
 32             (max(L2.pt1.y, L2.pt2.y) >= min(L1.pt1.y, L1.pt2.y)) &&
 33             (Multiply(L2.pt1, L1.pt2, L1.pt1) * Multiply(L1.pt2, L2.pt2, L1.pt1) >= 0) &&
 34             (Multiply(L1.pt1, L2.pt2, L2.pt1) * Multiply(L2.pt2, L1.pt2, L2.pt1) >= 0)
 35           );
 36 }
 37 /* 射线法判断点q与多边形polygon的位置关系，要求polygon为简单多边形，顶点逆时针排列
 38  如果点在多边形内： 返回0
 39  如果点在多边形边上： 返回1
 40  如果点在多边形外： 返回2
 41 */
 42 bool InPolygon(const Polygon& polygon, Point point)
 43 {
 44     int n = polygon.size();
 45     int count = 0;
 46     LineSegment line;
 47     line.pt1 = point;
 48     line.pt2.y = point.y;
 49     line.pt2.x = - inf;
 50 
 51     for( int i = 0; i < n; i++ )
 52     {
 53         LineSegment side;
 54         side.pt1 = polygon[i];
 55         side.pt2 = polygon[(i + 1) % n];
 56 
 57         if( IsOnline(point, side) )
 58         {
 59             return 1;
 60         }
 61 
 62         if( fabs(side.pt1.y - side.pt2.y) < ESP )
 63         {
 64             continue;
 65         }
 66 
 67         if( IsOnline(side.pt1, line) )
 68         {
 69             if( side.pt1.y > side.pt2.y ) count++;
 70         }
 71         else if( IsOnline(side.pt2, line) )
 72         {
 73             if( side.pt2.y > side.pt1.y ) count++;
 74         }
 75         else if( Intersect(line, side) )
 76         {
 77             count++;
 78         }
 79     }
 80 
 81     if ( count % 2 == 1 )
 82     {
 83         return 0;
 84     }
 85     else
 86     {
 87         return 2;
 88     }
 89 }
 90 void gao (int xx,int yy)
 91 {
 92     pp.clear();
 93     Point heart;
 94     heart.y = yy*2*o*0.5*sqrt(3);
 95     heart.x = yy*o+xx*2*o;
 96     Point p1;
 97     p1.x=heart.x;p1.y=heart.y+5.0;
 98 
 99     Point p2;
100     p2.x=heart.x+o;p2.y=heart.y+2.5;
101 
102     Point p3;
103     p3.x=heart.x+o;p3.y=heart.y-2.5;
104 
105     Point p4;
106     p4.x=heart.x;p4.y=heart.y-5.0;
107 
108     Point p5;
109     p5.x=heart.x-o;p5.y=heart.y-2.5;
110 
111     Point p6;
112     p6.x=heart.x-o;p6.y=heart.y+2.5;
113 
114     pp.push_back(p1);
115     pp.push_back(p2);
116     pp.push_back(p3);
117     pp.push_back(p4);
118     pp.push_back(p5);
119     pp.push_back(p6);
120 
121 }
122 int main()
123 {
124     //freopen("de.txt","r",stdin);
125     double xx,yy;
126     vector <pair <int ,int>> ans;
127     while (~scanf("%lf%lf",&xx,&yy)){
128         Point nowpt;
129         nowpt.x=xx,nowpt.y=yy;
130         bool f=false;
131         for (int i=-9;i<=10;++i){
132             for (int j=-9;j<=10;++j){
133                 if (!f){
134                     gao(i,j);
135                     if (InPolygon(pp,nowpt)==0){
136                         ans.push_back(make_pair(i,j));
137                         f=true;
138                         break;
139                     }
140                 }
141 
142             }
143         }
144 
145     }
146     int sz = ans.size();
147     for (int i=0;i<sz;++i){
148         printf("[%d,%d]",ans[i].first,ans[i].second);
149         if (i!=sz-1){
150             printf(", ");
151         }
152         else{
153             printf("n");
154         }
155 
156     }
157     return 0;
158 }