Moon Game
Description
Fat brother and Maze
are playing a kind of special (hentai) game in the clearly blue sky
which we can just consider as a kind of two-dimensional plane. Then Fat
brother starts to draw N starts in the sky which we can just consider
each as a point. After he draws these stars, he starts to sing the
famous song “The Moon Represents My Heart” to Maze.
You ask me how deeply I love you,
How much I love you?
My heart is true,
My love is true,
The moon represents my heart.
…
But
as Fat brother is a little bit stay-adorable(呆萌), he just consider that
the moon is a special kind of convex quadrilateral and starts to count
the number of different convex quadrilateral in the sky. As this number
is quiet large, he asks for your help.
Input
The first line of the date is an integer T, which is the number of the text cases.
Then T cases follow, each case contains an integer N describe the number of the points.
Then
N lines follow, Each line contains two integers describe the coordinate
of the point, you can assume that no two points lie in a same
coordinate and no three points lie in a same line. The coordinate of the
point is in the range[-10086,10086].
1 <= T <=100, 1 <= N <= 30
Output
For
each case, output the case number first, and then output the number of
different convex quadrilateral in the sky. Two convex quadrilaterals are
considered different if they lie in the different position in the sky.
Sample Input
2
4
0 0
100 0
0 100
100 100
4
0 0
100 0
0 100
10 10
Sample Output
Case 1: 1
Case 2: 0
这个题目是一道计算凸四边形个数的题目，只需要暴力枚举所有4边形判断即可。
判断的时候就是枚举对角线，只有有一种情况的两条对角线相交即为凸四边形。这里采用线段相交的模板来判断。
代码：
1 #include <iostream>
2 #include <cstdio>
3 #include <cstdlib>
4 #include <cstring>
5 #include <cmath>
6 #include <algorithm>
7 #include <set>
8 #include <map>
9 #include <queue>
10 #include <string>
11 #include <vector>
12 #define inf 0x3fffffff
13
14 using namespace std;
15
16 struct point
17 {
18 int x, y;
19 }p[35];
20
21 point a1, a2, a3, a4;
22
23 bool inter(point a, point b, point c, point d)
24 {
25 if(
26 min(a.x, b.x) > max(c.x, d.x) ||
27 min(a.y, b.y) > max(c.y, d.y) ||
28 min(c.x, d.x) > max(a.x, b.x) ||
29 min(c.y, d.y) > max(a.y, b.y)
30 )
31 return 0;
32 long long x1, x2, x3, x4;
33 x1 = (b.x - a.x) * (c.y - a.y) - (b.y - a.y) * (c.x - a.x);
34 x2 = (b.x - a.x) * (d.y - a.y) - (b.y - a.y) * (d.x - a.x);
35 x3 = (d.x - c.x) * (a.y - c.y) - (d.y - c.y) * (a.x - c.x);
36 x4 = (d.x - c.x) * (b.y - c.y) - (d.y - c.y) * (b.x - c.x);
37 return x1 * x2 < 0 && x3 * x4 < 0;
38 }
39
40 bool fun()
41 {
42 if (inter(a1, a2, a3, a4)) return 1;
43 if (inter(a1, a3, a2, a4)) return 1;
44 if (inter(a1, a4, a2, a3)) return 1;
45 return 0;
46 }
47
48 int main()
49 {
50 //freopen ("test.txt", "r", stdin);
51 int T;
52 scanf ("%d", &T);
53 for (int times = 1; times <= T; ++times)
54 {
55 int n;
56 scanf ("%d", &n);
57 for (int i = 0; i < n; ++i)
58 {
59 scanf ("%d%d", &p[i].x, &p[i].y);
60 }
61 int ans = 0;
62 for (int i = 0; i < n; ++i)
63 {
64 a1 = p[i];
65 for (int j = i + 1; j < n; ++j)
66 {
67 a2 = p[j];
68 for (int k = j + 1; k < n; ++k)
69 {
70 a3 = p[k];
71 for (int h = k + 1; h < n; ++h)
72 {
73 a4 = p[h];
74 if (fun()) ans++;
75 }
76 }
77 }
78 }
79 printf ("Case %d: %d\n", times, ans);
80 }
81 return 0;
82 }