ZOJ 3720 Magnet Darts (计算几何，概率，判点是否在多边形内)

浏览数：88 / 时间：2015年06月09日

http://acm.zju.edu.cn/onlinejudge/showProblem.do?problemCode=3720

题意：

在一个矩形区域投掷飞镖，因此飞镖只会落在整点上，投到每个点的得分是Ax+By。矩形区域里面有个多边形，如果飞镖投在多边形里面则得分，求最终的得分期望。

即：给定一个矩形内的所有整数点，判断这些点是否在一个多边形内

方法：

计算几何的判点是否在多边形内（几何模板），如果在，则令得分加(Ax+By)*以此点为中心边长为1的正方形面积

 1 void solve()
 2 {
 3     double ans = 0;
 4     for (int i = p.x; i <= q.x; i++)
 5     {
 6         for (int j = p.y; j <= q.y; j++)
 7         {
 8             if (PointInPolygon(Point2D(i, j), poly.v, poly.n))
 9                 ans += (a * i + b * j) * (min(i + 0.5, q.x) - max(i - 0.5, p.x)) * (min(j + 0.5, q.y) - max(j - 0.5, p.y));
10         }
11     }
12     printf("%.3f\n", ans / (q.x - p.x) / (q.y - p.y));
13 }

代码：

  1 // #pragma comment(linker, "/STACK:102400000,102400000")
  2 #include <cstdio>
  3 #include <iostream>
  4 #include <cstring>
  5 #include <string>
  6 #include <cmath>
  7 #include <set>
  8 #include <list>
  9 #include <map>
 10 #include <iterator>
 11 #include <cstdlib>
 12 #include <vector>
 13 #include <queue>
 14 #include <stack>
 15 #include <algorithm>
 16 #include <functional>
 17 using namespace std;
 18 typedef long long LL;
 19 #define ROUND(x) round(x)
 20 #define FLOOR(x) floor(x)
 21 #define CEIL(x) ceil(x)
 22 const int maxn = 0;
 23 const int maxm = 0;
 24 const int inf = 0x3f3f3f3f;
 25 const LL inf64 = 0x3f3f3f3f3f3f3f3fLL;
 26 // const double INF = 1e30;
 27 // const double eps = 1e-6;
 28 const int P[4] = {0, 0, -1, 1};
 29 const int Q[4] = {1, -1, 0, 0};
 30 const int PP[8] = { -1, -1, -1, 0, 0, 1, 1, 1};
 31 const int QQ[8] = { -1, 0, 1, -1, 1, -1, 0, 1};
 32 
 33 typedef double db;
 34 const double eps = 1e-7;
 35 const double PI = acos(-1.0);
 36 const double INF = 1e50;
 37 
 38 const int POLYGON_MAX_POINT = 1024;
 39 
 40 db dmin(db a, db b)
 41 {
 42     return a > b ? b : a;
 43 }
 44 db dmax(db a, db b)
 45 {
 46     return a > b ? a : b;
 47 }
 48 int sgn(db a)
 49 {
 50     return a < -eps ? -1 : a > eps;    //返回double型的符号
 51 }
 52 
 53 struct Point2D
 54 {
 55     db x, y;
 56     int id;
 57     Point2D(db _x = 0, db _y = 0): x(_x), y(_y) {}
 58     void input()
 59     {
 60         scanf("%lf%lf" , &x, &y);
 61     }
 62     void output()
 63     {
 64         printf("%.2f %.2f\n" , x, y);
 65     }
 66     //
 67     db len2()
 68     {
 69         return x * x + y * y;
 70     }
 71     //到原点距离
 72     double len()
 73     {
 74         return sqrt(len2());
 75     }
 76     //逆时针转90度
 77     Point2D rotLeft90()
 78     {
 79         return Point2D(-y, x);
 80     }
 81     //顺时针转90度
 82     Point2D rotRight90()
 83     {
 84         return Point2D(y, -x);
 85     }
 86     //绕原点逆时针旋转arc_u
 87     Point2D rot(double arc_u)
 88     {
 89         return Point2D( x * cos(arc_u) - y * sin(arc_u),
 90                         x * sin(arc_u) + y * cos(arc_u) );
 91     }
 92     //绕某点逆时针旋转arc_u
 93     Point2D rotByPoint(Point2D &center, db arc_u)
 94     {
 95         Point2D tmp( x - center.x, y - center.y );
 96         Point2D ans = tmp.rot(arc_u);
 97         ans = ans + center;
 98         return ans;
 99     }
100 
101     bool operator == (const Point2D &t) const
102     {
103         return sgn(x - t.x) == 0 && sgn(y - t.y) == 0;
104     }
105     bool operator < (const Point2D &t) const
106     {
107         if ( sgn(x - t.x) == 0 ) return y < t.y;
108         else return x < t.x;
109     }
110     Point2D operator + (const Point2D &t) const
111     {
112         return Point2D(x + t.x, y + t.y);
113     }
114 
115     Point2D operator - (const Point2D &t) const
116     {
117         return Point2D(x - t.x, y - t.y);
118     }
119 
120     Point2D operator * (const db &t)const
121     {
122         return Point2D( t * x, t * y );
123     }
124 
125     Point2D operator / (const db &t) const
126     {
127         return Point2D( x / t, y / t );
128     }
129     //点乘
130     db operator ^ (const Point2D &t) const
131     {
132         return x * t.x + y * t.y;
133     }
134     //叉乘
135     db operator * (const Point2D &t) const
136     {
137         return x * t.y - y * t.x;
138     }
139     //两点之间的角度(-PI , PI]
140     double rotArc(Point2D &t)
141     {
142         double perp_product = rotLeft90()^t;
143         double dot_product = (*this)^t;
144         if ( sgn(perp_product) == 0 && sgn(dot_product) == -1) return PI;
145         return sgn(perp_product) * acos( dot_product / len() / t.len() );
146     }
147     //标准化
148     Point2D normalize()
149     {
150         return Point2D(x / len(), y / len());
151     }
152 };
153 
154 struct POLYGON
155 {
156     Point2D v[POLYGON_MAX_POINT];
157     int n;
158 };
159 struct Segment2D
160 {
161     Point2D s , e;
162     Segment2D() {}
163     Segment2D( Point2D _s, Point2D _e ): s(_s), e(_e) {}
164 };
165 //0: 点在多边形外
166 //1: 点在多边形内
167 bool PonSegment2D(Point2D p, Segment2D seg)
168 {
169     return sgn((seg.s - p) * (p - seg.e)) == 0 && sgn((seg.s - p) ^ (p - seg.e)) >= 0;
170 }
171 
172 /**** 判断线段在内部相交***/
173 bool SegIntersect2D(Segment2D a, Segment2D b)
174 {
175     //必须在线段内相交
176     int d1 = sgn((a.e - a.s) * (b.s - a.s));
177     int d2 = sgn((a.e - a.s) * (b.e - a.s));
178     int d3 = sgn((b.e - b.s) * (a.s - b.s));
179     int d4 = sgn((b.e - b.s) * (a.e - b.s));
180     if (d1 * d2 < 0 && d3 * d4 < 0) return true;
181     return false;
182 }
183 
184 int PointInPolygon(Point2D p, Point2D poly[], int n)
185 {
186     Segment2D l, seg;
187     l.s = p;
188     l.e = p;
189     l.e.x = INF;//作射线
190     int i, cnt = 0;
191     Point2D p1, p2, p3, p4;
192     for (i = 0; i < n; i++)
193     {
194         seg.s = poly[i], seg.e = poly[(i + 1) % n];
195         if (PonSegment2D(p, seg)) return 2; //点在多边形上
196         p1 = seg.s, p2 = seg.e , p3 = poly[(i + 2) % n], p4 = poly[(i + 3) % n];
197         if (SegIntersect2D(l, seg) ||
198                 PonSegment2D(p2, l) && sgn((p2 - p1) * (p - p1))*sgn((p3 - p2) * (p - p2)) > 0 ||
199                 PonSegment2D(p2, l) && PonSegment2D(p3, l) &&
200                 sgn((p2 - p1) * (p - p1))*sgn((p4 - p3) * (p - p3)) > 0 )
201             cnt++;
202     }
203     if (cnt % 2 == 1) return 1; //点在多边形内
204     return 0;//点在多边形外
205 }
206 
207 POLYGON poly;
208 Point2D p, q;
209 db a, b;
210 void init()
211 {
212     //
213 }
214 void input()
215 {
216     scanf("%d%lf%lf", &poly.n, &a, &b);
217     for (int i = 0; i < poly.n; i++)
218     {
219         scanf("%lf%lf", &poly.v[i].x, &poly.v[i].y);
220     }
221 }
222 void debug()
223 {
224     //
225 }
226 void solve()
227 {
228     double ans = 0;
229     for (int i = p.x; i <= q.x; i++)
230     {
231         for (int j = p.y; j <= q.y; j++)
232         {
233             if (PointInPolygon(Point2D(i, j), poly.v, poly.n))
234                 ans += (a * i + b * j) * (min(i + 0.5, q.x) - max(i - 0.5, p.x)) * (min(j + 0.5, q.y) - max(j - 0.5, p.y));
235         }
236     }
237     printf("%.3f\n", ans / (q.x - p.x) / (q.y - p.y));
238 }
239 void output()
240 {
241     //
242 }
243 int main()
244 {
245     // std::ios_base::sync_with_stdio(false);
246     // #ifndef ONLINE_JUDGE
247     //     freopen("in.cpp", "r", stdin);
248     // #endif
249 
250     while (~scanf("%lf%lf%lf%lf", &p.x, &p.y, &q.x, &q.y))
251     {
252         init();
253         input();
254         solve();
255         output();
256     }
257     return 0;
258 }