POJ 2318(计算几何初步——叉积加2分)
呃。。。。很经典的题目，由于数据量较大，2分的话会比较省时间。
就是问你N个方块里面有多少个点。
so easy！
View Code
1 #include<iostream>
2 using namespace std;
3 int px,py;
4 struct LINE
5 {
6 int Ux, Uy, Lx, Ly;
7 }line[5002];
8 int sum[5001];
9 int n, m;
10
11 inline int cross(int x1, int y1, int x2, int y2)
12 {
13 return x1*y2-y1*x2;
14 }
15
16 inline bool in_box(int cnt)
17 {
18 if (cross(px-line[cnt].Lx,py-line[cnt].Ly,line[cnt].Ux-line[cnt].Lx,line[cnt].Uy-line[cnt].Ly)>0)
19 return true;
20 else
21 return false;
22 }
23
24 int main()
25 {
26 while (scanf("%d",&n) && n)
27 {
28 scanf("%d%d%d%d%d",&m,&line[0].Ux,&line[0].Uy,&line[n+1].Lx,&line[n+1].Ly);
29 line[0].Lx = line[0].Ux;
30 line[0].Ly = line[n+1].Ly;
31 line[n+1].Ux = line[n+1].Lx;
32 line[n+1].Uy = line[0].Uy;
33 sum[0] = 0;
34 for (int i(1); i<=n; ++i)
35 {
36 scanf("%d%d",&line[i].Ux,&line[i].Lx);
37 line[i].Uy = line[0].Uy;
38 line[i].Ly = line[n+1].Ly;
39 sum[i] = 0;
40 }
41 for (int k(1); k<=m; ++k)
42 {
43 scanf("%d%d",&px,&py);
44 int l(0);
45 int r(n+1);
46 while (l <= r)
47 {
48 int mid = (l + r) >> 1;
49 if (in_box(mid))
50 {
51 if (l+1 == r)
52 {
53 sum[mid]++;
54 break;
55 }
56 l = mid;
57 }
58 else
59 {
60 if (l+1 == r)
61 {
62 sum[mid]++;
63 break;
64 }
65 r = mid;
66 }
67 }
68 }
69 for (int i(0); i<=n; ++i)
70 {
71 printf("%d: %d\n",i,sum[i]);
72 }
73 printf("\n");
74 }
75 return 0;
76 }
