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

題意：

在一個矩形區域投擲飛鏢，因此飛鏢只會落在整點上，投到每個點的得分是Ax+By。矩形區域里面有個多邊形，如果飛鏢投在多邊形里面則得分，求最終的得分期望。

即：給定一個矩形內的所有整數點，判斷這些點是否在一個多邊形內

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方法：

計算幾何的判點是否在多邊形內（幾何模板），如果在，則令得分加(Ax+By)*以此點為中心邊長為1的正方形面積

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 }

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代碼：

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型的符號 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 //到原點距離 72 double len() 73 { 74 return sqrt(len2()); 75 } 76 //逆時針轉90度 77 Point2D rotLeft90() 78 { 79 return Point2D(-y, x); 80 } 81 //順時針轉90度 82 Point2D rotRight90() 83 { 84 return Point2D(y, -x); 85 } 86 //繞原點逆時針旋轉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 //繞某點逆時針旋轉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 //點乘 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 //兩點之間的角度(-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 //標準化 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: 點在多邊形外 166 //1: 點在多邊形內 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 /**** 判斷線段在內部相交***/ 173 bool SegIntersect2D(Segment2D a, Segment2D b) 174 { 175 //必須在線段內相交 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;//作射線 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; //點在多邊形上 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; //點在多邊形內 204 return 0;//點在多邊形外 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 } View Code

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轉載于:https://www.cnblogs.com/xysmlx/p/3870091.html

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