卡了几天的破题，对于hdu的那份数据，这就一神题。。
借助极角排序，枚举以每一个点进行极角排序，然后构造两条扫描线，一个上面一个下面，两条同时走，把上线和下线的点以及上线左边的点分别统计出来，如下图
样例3：
假如现在以d为p[0],那么所有可能结果一定是他与其他点的连线所分割的平面，那么首先以de为上线，下线的角度为上线+pi，两条线始终维护着这样的关系。de的下一个点为f，di的下一个点为c ，比较一下两者需要转的角度，选取较小角度转，注意一下相同的时候的处理。
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 }
poj2280Amphiphilic Carbon Molecules(极角排序)
时间： 2024-10-21 10:52:26