hlg1306再遇攻击--射线法判断点是否在多边形内部

 1 #include <iostream>
 2 #include <cstdio>
 3 #include <cstring>
 4 #include <algorithm>
 5 using namespace std;
 6 
 7 const int maxn = 55;
 8 
 9 int min(int a, int b)
10 {
11     return a > b ? b : a;
12 }
13 int max(int a, int b)
14 {
15     return a > b ? a : b;
16 }
17 
18 struct point
19 {
20     int x;
21     int y;
22 }poin[maxn];
23 
24 int cross(point p0, point p1, point p2)
25 {
26     return (p1.x - p0.x)*(p2.y - p0.y) - (p2.x - p0.x)*(p1.y - p0.y);
27 }
28 
29 bool check_online(point p1, point p2, point Q)
30 {
31     if(cross(p1, Q, p2) == 0 &&
32        Q.x >= min(p1.x, p2.x) && Q.x <= max(p1.x, p2.x) &&
33        Q.y >= min(p1.y, p2.y) && Q.y <= max(p1.y, p2.y) )
34         return true;
35     return false;
36 }
37 
38 bool check_two_line(point a,point b, point c, point d)
39 {
40     int h, i, j, k;
41     h = cross(a, b, c);
42     i = cross(a, b, d);
43     j = cross(c, d, a);
44     k = cross(c, d, b);
45     return h * i <= 0 && j * k <= 0;
46 }
47 
48 int main()
49 {
50     int n;
51     while(scanf("%d",&n) && n)
52     {
53         point Q, P, p1, p2;
54         scanf("%d %d",&Q.x, &Q.y);
55         P.x = 101, P.y = Q.y;
56         for(int i = 0; i < n; i++)
57             scanf("%d %d",&poin[i].x, &poin[i].y);
58         bool flag = false;
59         int cnt = 0;
60         for(int i = 0; i < n; i++)
61         {
62             p1 = poin[i], p2 = poin[(i+1)%n];
63             if(check_online(p1, p2, Q))
64             {
65                 flag = true;
66                 break;
67             }
68             if(p1.y == p2.y)
69                 continue;
70             if(Q.y > min(p1.y, p2.y) && Q.y <= max(p1.y, p2.y))
71             {
72                 if(check_two_line(Q, P, p1, p2))
73                     cnt++;
74             }
75         }
76         //printf("%d\n",cnt);
77         if(flag)
78         {
79             printf("Yes\n");
80             continue;
81         }
82         if(cnt % 2 == 1)
83             printf("Yes\n");
84         else
85             printf("No\n");
86     }
87     return 0;
88 }