POJ2398+几何+排序

题意：

给定一些线段的两端和一些点。

问某些区域出现的点的个数的次数。。。

题目给定的线段未排序！！！

View Code

 1 /*

 2 几何+排序

 3 */

 4 #include<stdio.h>

 5 #include<string.h>

 6 #include<stdlib.h>

 7 #include<algorithm>

 8 #include<iostream>

 9 using namespace std;

10 const int maxn = 1005;

11 const double eps = 1e-8;

12 struct point {

13     double x,y;

14 };

15 struct Line {

16     point u,d;

17 }line[ maxn ];

18 int cnt[ maxn ],vis[ maxn ];

19 double xmult( point sp,point ep,point op ){

20     return ( sp.x-op.x )*( ep.y-op.y )-( sp.y-op.y )*( ep.x-op.x );

21 }

22 int cmp( Line a,Line b ){

23     return ( a.u.x+a.d.x )<( b.u.x+b.d.x );

24 }

25 int main(){

26     int n,m;

27     double x1,x2,y1,y2;

28     while( scanf("%d",&n),n ){

29         scanf("%d%lf%lf%lf%lf",&m,&x1,&y1,&x2,&y2);

30         line[ 0 ].u.x = x1,line[ 0 ].u.y = y1;

31         line[ 0 ].d.x = x1,line[ 0 ].d.y = y2;

32         for( int i=1;i<=n;i++ ){

33             scanf("%lf%lf",&line[ i ].u.x,&line[ i ].d.x);

34             line[ i ].u.y = y1;

35             line[ i ].d.y = y2;

36         }

37         sort( line+1,line+n+1,cmp );

38         line[ n+1 ].u.x = x2,line[ n+1 ].u.y = y1;

39         line[ n+1 ].d.x = x2,line[ n+1 ].d.y = y2;

40         memset( cnt,0,sizeof( cnt ) );

41         memset( vis,0,sizeof( vis ) );

42         for( int num=0;num<m;num++ ){

43             point toy;

44             scanf("%lf%lf",&toy.x,&toy.y);

45             for( int j=0;j<=n;j++ ){

46                 if( xmult( toy,line[ j ].d,line[ j ].u )<-eps&&xmult( toy,line[ j+1 ].d,line[ j+1 ].u )>eps ){

47                     cnt[ j ]++;

48                 }

49             }

50         }

51         for( int i=0;i<=n;i++ ){

52             if( cnt[ i ]!=0 )

53                 vis[ cnt[i] ]++;

54         }

55         printf("Box\n");

56         for( int i=1;i<maxn;i++ ){

57             if( vis[ i ]!=0 )

58                 printf("%d: %d\n",i,vis[ i ]);

59         }

60     }

61     return 0;

62 }