题目大意就是说两个人掷飞镖，飞镖在所给定的图形内就记一分，现在给定N个图形（圆、三角形和矩形），问每一次比赛（没人分别掷三次）谁赢。
1 #include <map>
2 #include <set>
3 #include <stack>
4 #include <queue>
5 #include <cmath>
6 #include <ctime>
7 #include <vector>
8 #include <cstdio>
9 #include <cctype>
10 #include <cstring>
11 #include <cstdlib>
12 #include <iostream>
13 #include <algorithm>
14 using namespace std;
15 #define eps 1e-12
16 #define MAXN 55
17 #define INF 1e30
18 #define mem0(a) memset(a,0, sizeof(a))
19 #define mem1(a) memset(a,-1,sizeof(a))
20 double MAX(double a, double b) {return a > b ? a : b;}
21 double MIn(double a, double b) {return a < b ? a : b;}
22 typedef long long LL;
23 /****************************************计算几何头文件**************************************************/
24 struct Point{
25 double x,y;
26 Point(double x=0, double y=0):x(x),y(y){}
27 };
28
29 struct Polygon
30 {
31 Point p[MAXN];
32 int Size;
33 };
34
35 struct Circle
36 {
37 Point o;
38 double r;
39 Circle(){}
40 Circle(Point _o, double _r):o(_o),r(_r){}
41 };
42
43 Point operator + (Point A, Point B) {return Point(A.x+B.x, A.y+B.y);}
44
45 Point operator - (Point A, Point B) {return Point(A.x-B.x, A.y-B.y);}
46
47 Point operator * (Point A, double p) {return Point(A.x*p, A.y*p);}
48
49 Point operator / (Point A, double p) {return Point(A.x/p, A.y/p);}
50
51 int dcmp(double x) {
52 if(fabs(x) < eps) return 0;
53 else return x < 0 ? -1 : 1;
54 }
55
56 bool operator == (const Point &A, const Point &B) {
57 return dcmp(A.x-B.x) == 0 && dcmp(A.y-B.y) == 0;
58 }
59
60 double Dot(Point A, Point B) { return A.x*B.x + B.y*B.y;} //点积
61
62 double Length(Point A) { return sqrt(Dot(A,A));} //向量长度
63
64 double Angle(Point A, Point B) {return acos(Dot(A,B) / Length(A) / Length(B));}//向量夹角
65
66 double cross(Point A, Point B) {return A.x*B.y - A.y*B.x;}
67
68 bool crossed(Point a, Point b, Point c, Point d)//线段ab和cd是否相交
69 {
70 if(cross(a-c, d-c)*cross(b-c, d-c)<=0 && cross(c-a, b-a)*cross(d-a, b-a)<=0)
71 {
72 return true;
73 }
74 return false;
75 }
76
77 bool isPointOnSegent(Point p, Point s, Point e)//判断点是否在线段se上
78 {
79 double d = (p.x-s.x) * (e.x-p.x);
80 double a = (p.y-s.y) / (p.x-s.x);
81 double b = (e.y-p.y) / (e.x-p.x);
82 if(dcmp(d)==1 && dcmp(a-b)==0)return true;
83 return false;
84 }
85
86 int isPointInPolygon(Point p, Polygon poly)//判断点是否在多边形以内
87 {
88 int w = 0;
89 int n = poly.Size;
90 for(int i=0;i<n;i++)
91 {
92 if(isPointOnSegent(p, poly.p[i], poly.p[(i+1)%n])) return 1;//点在边上
93 int k = dcmp(cross(poly.p[(i+1)%n]-poly.p[i], p-poly.p[i]));
94 int d1 = dcmp(poly.p[i].y - p.y);
95 int d2 = dcmp(poly.p[(i+1)%n].y - p.y);
96 if(k > 0 && d1 <= 0 && d2 > 0) w++;
97 if(k < 0 && d2 <= 0 && d1 > 0) w--;
98 }
99 if(w != 0) return 1;
100 return 0;
101 }
102
103 /****************************************************************************************************/
104 struct R
105 {
106 Point a, b;
107 R(){}
108 R(Point _a, Point _b)
109 {
110 a = _a;
111 b = _b;
112 }
113 }r[1001];
114 struct T
115 {
116 Point a, b, c;
117 T(){}
118 T(Point _a, Point _b, Point _c) {
119 a = _a; b = _b; c = _c;
120 }
121 }t[1001];
122 Circle c[1001];
123 int S, N;
124 int cntC = 0, cntT = 0, cntR = 0;
125
126 int calc(double x, double y)//计算点(x, y)在图中的多少个图形内部
127 {
128 Point p = Point(x, y);
129 int ans = 0;
130 for(int i=0;i<cntC;i++)
131 {
132 if(Length(p-c[i].o) <= c[i].r) ans ++;
133 }
134 for(int i=0;i<cntT;i++)
135 {
136 if( cross(t[i].c-t[i].a, p-t[i].a)*cross(t[i].b-t[i].a, p-t[i].a)<=0
137 && cross(t[i].a-t[i].b, p-t[i].b)*cross(t[i].c-t[i].b, p-t[i].b)<=0 ) ans ++;
138 }
139 for(int i=0;i<cntR;i++)
140 {
141 if(x>=r[i].a.x&&x<=r[i].b.x && y>=r[i].a.y&&y<=r[i].b.y) ans ++;
142 }
143 return ans;
144 }
145
146 int main()
147 {
148 char ch;double x, y, rr;
149 while(~scanf("%d%*c", &S))
150 {
151 cntT = cntC = cntR = 0;
152 for(int i=0;i<S;i++)
153 {
154 scanf("%c", &ch);
155 if(ch == ‘C‘)
156 {
157 scanf("%lf %lf %lf%*c", &x, &y, &rr);
158 c[cntC++] = Circle(Point(x, y), rr);
159 continue;
160 }
161 else if(ch == ‘T‘)
162 {
163 Point aa[3];
164 for(int j=0;j<3;j++)
165 {
166 scanf("%lf%*c%lf%*c", &x, &y);
167 aa[j] = Point(x, y);
168 }
169 t[cntT++] = T(aa[0],aa[1],aa[2]);
170 }
171 else
172 {
173 Point aa[2];
174 for(int j=0;j<2;j++)
175 {
176 scanf("%lf%*c%lf%*c", &x, &y);
177 aa[j] = Point(x, y);
178 }
179 r[cntR++] = R(aa[0], aa[1]);
180 }
181 }
182 scanf("%d", &N);
183 for(int i=0;i<N;i++)
184 {
185 int cntA = 0, cntB = 0;
186 for(int j=0;j<3;j++)
187 {
188 scanf("%lf %lf", &x, &y);
189 cntA += calc(x, y);
190 }
191 for(int j=0;j<3;j++)
192 {
193 scanf("%lf %lf", &x, &y);
194 cntB += calc(x, y);
195 }
196 // printf("%d %d\n", cntA, cntB);
197 printf("%s\n", cntA > cntB ? "Bob" : (cntA == cntB ? "Tied" : "Hannah"));
198 }
199 }
200 return 0;
201 }
时间： 2024-10-09 19:01:17