POJ3714+最近点对

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 1 #include<stdio.h>

 2 #include<string.h>

 3 #include<stdlib.h>

 4 #include<math.h>

 5 #include<algorithm>

 6 using namespace std;

 7 const double eps = 1e-8;

 8 const double inf = 9999999999.0;

 9 const int maxn =  100005;

10 struct Point{

11     double x,y;

12     int flag;

13 };

14 Point pnt[ maxn<<1 ],temp[ maxn<<1 ];

15 double dis( Point a,Point b ){

16     return sqrt( (a.x-b.x)*(a.x-b.x)+(a.y-b.y)*(a.y-b.y) );

17 }

18 int cmpxy( Point a,Point b ){

19     if( a.x!=b.x )

20         return a.x<b.x;

21     else

22         return a.y<b.y;

23 }

24 int cmpx( Point a,Point b ){

25     return a.x<b.x;

26 }

27 int cmpy( Point a,Point b ){

28     return a.y<b.y;

29 }

30 double solve( int L,int R ){

31     if( L==R )

32         return inf;

33     if( L+1==R ){

34         if( pnt[L].flag==pnt[R].flag )

35             return inf;

36         else

37             return dis( pnt[L],pnt[R] );

38     }

39     int mid = (L+R)/2;

40     double res,Ldis,Rdis;

41     Ldis = solve( L,mid );

42     Rdis = solve( mid+1,R );

43     res = min( Ldis,Rdis );

44     int cnt = 0;

45     for( int i=L;i<=R;i++ ){

46         if( fabs(pnt[i].x-pnt[mid].x)<=res ){

47             temp[cnt++] = pnt[i];

48         }

49     }

50     sort( temp,temp+cnt,cmpy );

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

52         for( int j=i+1;j<cnt;j++ ){

53             if( fabs( pnt[i].y-pnt[j].y )>res ) break;

54             if( pnt[i].flag==pnt[j].flag ) continue;

55             res = min( res,dis(pnt[i],pnt[j]) );

56         }

57     }

58     return res;

59 }

60 

61 int main(){

62     int ca;

63     scanf("%d",&ca);

64     while( ca-- ){

65         int n;

66         scanf("%d",&n);

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

68             scanf("%lf%lf",&pnt[i].x,&pnt[i].y);

69             pnt[i].flag = 1;

70         }

71         for( int i=n;i<2*n;i++ ){

72             scanf("%lf%lf",&pnt[i].x,&pnt[i].y);

73             pnt[i].flag = 2;

74         }

75         sort( pnt,pnt+2*n,cmpxy );

76         double Ans = solve( 0,2*n-1 );

77         printf("%.3lf\n",Ans);

78     }

79     return 0;

80 }

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