zoj1081 Points Within

1. //zoj1081 Points Within
2. //Accepted 1081 C++ 00:00.00 396K
3. //(?)处有问题，不过还是AC了
4. #include <stdio.h>
5. #define MAXN 100
6. #define INF 1e10
7. struct point { double x,y;};
8. int n;
9. double xmult(double x1,double y1,double x2,double y2){return x1*y2-x2*y1;}
10. double xmult3(point p1,point p2,point p0)
11. {return (p1.x-p0.x)*(p2.y-p0.y)-(p2.x-p0.x)*(p1.y-p0.y);}
12. //判点p1、p2在线段p3p4同侧(不包括在直线线上)
13. bool same_side(point p1,point p2,point p3,point p4)
14. {
15.        double dx,dy,dx1,dy1,dx2,dy2;
16.        dx = p4.x-p3.x, dy = p4.y-p3.y;
17.        dx1 = p1.x-p3.x, dy1 = p1.y-p3.y;
18.        dx2 = p2.x-p4.x, dy2 = p2.y-p4.y;
19.        return xmult(dx,dy,dx1,dy1)*xmult(dx,dy,dx2,dy2)>0.0;
20. }
21. //判p1p2与p3p4相交(包括端点)
22. bool intersect(point p1,point p2,point p3,point p4)
23. {
24.        return (!same_side(p1,p2,p3,p4) && !same_side(p3,p4,p1,p2));
25. }
26. //判点q在线段p1p2上(包括端点)
27. bool dot_online(point q,point p1,point p2)
28. {
29.        return xmult3(q,p1,p2)==0.0 &&
30.               (q.x-p1.x)*(q.x-p2.x)<=0.0 &&
31.               (q.y-p1.y)*(q.y-p2.y)<=0.0;
32. }
33. //判点q在多边形P(p[0,1,...,n-1]为P的邻接顶点)内(包括在边上)，射线穿刺、奇偶校验法
34. /* Special case
35.   y|       __
36.    |      /  /
37.    |     / .__/____//______qs
38.    |    /   q        /
39. ___|____/____________/____x
40.    |     /__________/
41. (a) q->qs overlap the edge of P
42.  
43.   y|       __
44.    |      /  /
45.    |     /    /_.__//
46.    |    /        q   /
47. ___|____/____________/____x
48.    |     /__________/
49. (b) q is on the edge of P (true)
50.  
51.   y|       __
52.    |      /._/_____________qs
53.    |     /  q /____//
54.    |    /            /
55. ___|____/____________/____x
56.    |     /__________/
57. (c) q->qs cross the vertex of P
58.  
59.   y|       __
60.    |      /  /
61.    |     /    /____//
62.    |    /  q.________/______qs
63. ___|____/____________/____x
64.    |     /__________/
65. (d) q-qs cross the vertex of P
66.  
67. Sample Input
68. 9 5
69.  
70. 1 1
71. 4 1
72. 6 3
73. 5 5
74. 4 4
75. 2 4
76. 1 6
77. -1 6
78. -3 3
79.  
80. 1 4(a)
81. 3 4(b)
82. 1 5(c)
83. 1 3(d)
84. -5 3(d)
85. 0
87. Sample Output
88. Within
89. Within
90. Within
91. Within
92. Outside
93. */
94. bool inside_polygon(point q,point *p,int n)
95. {
96.        point qs=q;
97.        qs.x = INF;//construct radial q->qs
98.        int cnt = 0;
99.        for (int i=0; i<n; ++i){
100.               int j=(i+1)%n,k=(i+2)%n;
101.               if (dot_online(q,p[i],p[j])) return true;//case (b)
102.               if (intersect(q,qs,p[j],p[k])) cnt++;//case (c) and common
103. //(?) if (dot_online(p[j],q,qs) && !same_side(p[i],p[k],q,qs)) cnt--;//case (d)
104. //(?) if (dot_online(p[j],q,qs) && dot_online(p[k],q,qs)) cnt--;//case (a)
105.        }
106.        printf ("%d",cnt);
107.        return cnt%2!=0;
108. }
109. void solve()
110. {
111.        int m;
112.        point q,p[MAXN];
113.        scanf ("%d",&m);
114.        for (int i=0; i<n; ++i) scanf ("%lf %lf",&p[i].x,&p[i].y);
115.        while (m--){
116.               scanf ("%lf %lf",&q.x,&q.y);
117.               if (inside_polygon(q,p,n))
118.                      printf ("Within/n");
119.               else printf ("Outside/n");
120.        }
121. }
122. int main()
123. {
124. #ifdef ONLINE_JUDGE
125. #else
126.        freopen("1081.txt","r",stdin);
127. #endif
128.        int t=0;
129.        while (scanf("%d",&n)!=EOF && n){
130.               if (t) printf ("/n");
131.               printf ("Problem %d:/n",++t);
132.               solve();
133.        }
134. #ifdef ONLINE_JUDGE
135. #else
136.        fclose(stdin);
137. #endif
138.        return 0;
139. }