poj2280Amphiphilic Carbon Molecules(极角排序)

  1 #include <iostream>
  2 #include<cstdio>
  3 #include<cstring>
  4 #include<algorithm>
  5 #include<stdlib.h>
  6 #include<vector>
  7 #include<cmath>
  8 #include<queue>
  9 #include<set>
 10 using namespace std;
 11 #define N 1010
 12 #define LL long long
 13 #define INF 0xfffffff
 14 const double eps = 1e-8;
 15 const double pi = acos(-1.0);
 16 const double inf = ~0u>>2;
 17 struct point
 18 {
 19     double x,y;
 20     double a;
 21     int flag;
 22     point(double x=0,double y = 0):x(x),y(y) {}
 23 } p[N],q[N];
 24 int tx[N],ty[N];
 25 typedef point pointt;
 26 point operator -(point a,point b)
 27 {
 28     return point(a.x-b.x,a.y-b.y);
 29 }
 30 int dcmp(double x)
 31 {
 32     if(fabs(x)<eps) return 0;
 33     return x<0?-1:1;
 34 }
 35 double cross(point a,point b)
 36 {
 37     return a.x*b.y-a.y*b.x;
 38 }
 39 double mul(point a,point b,point c)
 40 {
 41     return cross(b-a,c-a);
 42 }
 43 double dis(point a)
 44 {
 45     return sqrt(a.x*a.x+a.y*a.y);
 46 }
 47 int dot_online_in(point p,point l1,point l2)
 48 {
 49     return !dcmp(mul(p,l1,l2))&&(l1.x-p.x)*(l2.x-p.x)<eps&&(l1.y-p.y)*(l2.y-p.y)<eps;
 50 }
 51 double dot(point a,point b)
 52 {
 53     return a.x*b.x+a.y*b.y;
 54 }
 55 double Cal_Angle(point p1,point p2)
 56 {
 57     if(p1.x == p2.x && p1.y == p2.y)
 58         return -100.0;
 59     point v;
 60     v.x = p2.x - p1.x;
 61     v.y = p2.y - p1.y;
 62     if(p1.y <= p2.y)
 63         return acos(v.x/sqrt(dot(v,v)));
 64     return 2.0*pi - acos(v.x/sqrt(dot(v,v)));
 65 }
 66 
 67 void Cal_Angle(point pp,point p[],int n)
 68 {
 69     for(int i = 0; i < n; ++i)
 70     {
 71         p[i].a = Cal_Angle(pp,p[i]);
 72     }
 73 }
 74 bool cmp_angle(point p1,point p2)
 75 {
 76     return p1.a < p2.a;
 77 }
 78 int main()
 79 {
 80     int i,j,n;
 81     while(scanf("%d",&n)&&n)
 82     {
 83         int a =0,b =0 ;
 84         for(i = 0; i < n; i++)
 85         {
 86             scanf("%lf%lf%d",&p[i].x,&p[i].y,&p[i].flag);
 87             q[i] = p[i];
 88             if(p[i].flag) a++;
 89             else b++;
 90         }
 91         if(n<3)
 92         {
 93             printf("%d\n",n);
 94             continue;
 95         }
 96         int ans = max(a,b);
 97         for(i = 0; i < n; i++)
 98         {
 99             swap(q[i],q[0]);
100             for(j = 0 ; j < n; j++)
101                 p[j] = q[j];
102             Cal_Angle(p[0],p,n);
103             sort(p+1,p+n,cmp_angle);
104             for(j = n-1; j > 0 ; j--)
105                 p[j].a-=p[1].a;
106             double ani = p[1].a,anj = pi+ani;
107             int on_red = 0,on_blue = 0,under_red = 0, under_blue = 0,red = 0,blue = 0;
108             int g = 1;
109             while(dcmp(p[g].a-ani)==0&&g<n)
110             {
111                 p[g++].flag?on_red++:on_blue++;
112             }
113             int tk = n;
114             for(j = g; j < n; j++)
115             {
116                 if(dcmp(p[j].a-anj)>0)
117                 {
118                     tk = j;
119                     break;
120                 }
121                 if(dcmp(p[j].a-anj)==0)
122                 {
123                     while(dcmp(p[j].a-anj)==0&&j<n)
124                     {
125                         p[j++].flag?under_red++:under_blue++;
126                     }
127                     tk = j;
128                     break;
129                 }
130                 p[j].flag?red++:blue++;
131             }
132             int ta = 0,tb = 0;
133             p[0].flag?ta++:tb++;
134 
135             int k1 = red+on_red+under_red+1+b-blue-tb;
136             int k2 = blue+on_blue+under_blue+1+a-red-ta;
137 
138             ans = max(ans,max(k1,k2));
139             if(g==n) continue;
140 
141             while(tk<n)
142             {
143                 red+=under_red,blue+=under_blue;
144                 on_red = on_blue = under_blue = under_red = 0;
145 
146                 if(tk==n||p[tk].a-anj>p[g].a-ani)
147                 {
148                     ani = p[g].a;
149                     anj = ani+pi;
150                     while(dcmp(p[g].a-ani)==0&&g<n)
151                     {
152                         p[g].flag?on_red++:on_blue++;
153                         p[g++].flag?red--:blue--;
154                     }
155                 }
156                 else
157                 {
158                     anj = p[tk].a;
159                     ani = anj-pi;
160                     while(dcmp(p[tk].a-anj)==0&&tk<n)
161                     {
162                         p[tk++].flag?under_red++:under_blue++;
163                     }
164                     while(dcmp(p[g].a-ani)==0&&g<n)
165                     {
166                         p[g].flag?on_red++:on_blue++;
167                         p[g++].flag?red--:blue--;
168                     }
169                 }
170 
171                 int k1 = red+on_red+under_red+1+b-blue-tb;
172                 int k2 = blue+on_blue+under_blue+1+a-red-ta;
173                 ans = max(ans,max(k1,k2));
174             }
175         }
176         printf("%d\n",ans);
177     }
178     return 0;
179 }