POJ 2398 Toy Storage

这道题和POJ 2318几乎是一样的。

区别就是输入中坐标不给排序了，=_=||

输出变成了，有多少个区域中有t个点。

 1 #include <cstdio>

 2 #include <cmath>

 3 #include <cstring>

 4 #include <algorithm>

 5 using namespace std;

 6 

 7 struct Point

 8 {

 9     int x, y;

10     Point(int x=0, int y=0):x(x), y(y) {}

11 };

12 typedef Point Vector;

13 

14 Point read_point()

15 {

16     int x, y;

17     scanf("%d%d", &x, &y);

18     return Point(x, y);

19 }

20 

21 Point operator + (const Point& A, const Point& B)

22 { return Point(A.x+B.x, A.y+B.y); }

23 

24 Point operator - (const Point& A, const Point& B)

25 { return Point(A.x-B.x, A.y-B.y); }

26 

27 int Cross(const Point& A, const Point& B)

28 { return A.x*B.y - A.y*B.x; }

29 

30 const int maxn = 5000 + 10;

31 int up[maxn], down[maxn], ans[maxn];

32 int n, m;

33 Point A0, B0;

34 

35 int binary_search(const Point& P)

36 {

37     int L = 0, R = n;

38     while(L < R)

39     {

40         int mid = L + (R - L + 1) / 2;

41         Point A(down[mid], B0.y), B(up[mid], A0.y);

42         Vector v1 = B - A;

43         Vector v2 = P - A;

44         if(Cross(v1, v2) < 0) L = mid;

45         else R = mid - 1;

46     }

47     return L;

48 }

49 

50 int main()

51 {

52     //freopen("in.txt", "r", stdin);

53 

54     while(scanf("%d", &n) == 1 && n)

55     {

56         memset(ans, 0, sizeof(ans));

57 

58         scanf("%d", &m); A0 = read_point(); B0 = read_point();

59         for(int i = 1; i <= n; ++i) scanf("%d%d", &up[i], &down[i]);

60         sort(up + 1, up + 1 + n); sort(down + 1, down + 1 + n);

61         for(int i = 0; i < m; ++i)

62         {

63             Point P; P = read_point();

64             int pos = binary_search(P);

65             ans[pos]++;

66         }

67         sort(ans, ans + n + 1);

68 

69         puts("Box");

70         int i;

71         for(i = 0; i <= n; ++i) if(ans[i]) break;

72         while(i <= n)

73         {

74             int cnt = 1;

75             int num = ans[i];

76             while(i <= n && ans[i+1] == ans[i]) { i++; cnt++; }

77             i++;

78             printf("%d: %d\n", num, cnt);

79         }

80     }

81 

82     return 0;

83 }

代码君