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 ¢er, 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 }
ZOJ 3720 Magnet Darts (计算几何，概率，判点是否在多边形内),布布扣,bubuko.com
时间： 2024-10-15 10:29:00