ZOJ 1426 Counting Rectangles

枚举两条竖线之间同时与这两条竖线相交的横线的个数cnt，这两条竖线之间能组成矩形的个数为：从cnt中任选两条横线的方案个数。

将每两条竖线之间能组成矩形的个数加起来就是结果。

 

 1 #include <cstdio>

 2 #include <cstring>

 3 #include <cstdlib>

 4 #include <algorithm>

 5 #include <vector>

 6 

 7 using namespace std;

 8 

 9 const int MAXN = 110;

10 

11 struct MyLine

12 {

13     int st;

14     int ed;

15     int th;

16     MyLine() {}

17     MyLine( int _st, int _ed, int _th ): st(_st), ed(_ed), th(_th) { }

18 };

19 

20 vector<MyLine> hori;

21 vector<MyLine> vert;

22 bool ok[MAXN][MAXN];

23 

24 int solved()

25 {

26     int ans = 0;

27     int lenH = hori.size();

28     int lenV = vert.size();

29 

30     memset( ok, false, sizeof(ok) );

31 

32     for ( int i = 0; i < lenH; ++i )

33     {

34         for ( int j = 0; j < lenV; ++j )

35             if ( vert[j].th >= hori[i].st && vert[j].th <= hori[i].ed

36                 && hori[i].th >= vert[j].st && hori[i].th <= vert[j].ed ) ok[i][j] = true;

37     }

38 

39 

40     for ( int i = 0; i < lenH; ++i )

41     for ( int j = i + 1; j < lenH; ++j )

42     {

43         int cnt = 0;

44         for ( int k = 0; k < lenV; ++k )

45         if ( ok[i][k] && ok[j][k] ) ++cnt;

46         ans += cnt * (cnt - 1) / 2;

47     }

48 

49     return ans;

50 }

51 

52 int main()

53 {

54    // freopen( "input.txt", "r", stdin );

55     int T;

56     scanf( "%d", &T );

57     while ( T-- )

58     {

59         int N;

60         scanf( "%d", &N );

61         int x1, y1, x2, y2;

62         vert.clear();

63         hori.clear();

64         for ( int i = 0; i < N; ++i )

65         {

66             scanf( "%d%d%d%d", &x1, &y1, &x2, &y2 );

67             if ( x1 == x2 )

68             {

69                 if ( y1 > y2 ) swap( y1, y2 );

70                 hori.push_back( MyLine( y1, y2, x1 ) );

71             }

72             else if ( y1 == y2 )

73             {

74                 if ( x1 > x2 ) swap( x1, x2 );

75                 vert.push_back( MyLine( x1, x2, y1 ) );

76             }

77         }

78 

79         printf( "%d\n", solved() );

80     }

81     return 0;

82 }