二分+叉积判断方向 poj 2318

Posted 2020-08-07 blues

 1 // 题意：问你每个区域有多少个点
 2 // 思路：数据小可以直接暴力
 3 // 也可以二分区间
 4 
 5 #include <cstdio>
 6 #include <cstring>
 7 #include <iostream>
 8 #include <algorithm>
 9 #include <map>
10 #include <set>
11 #include <queue>
12 #include <stdlib.h>
13 #include <cmath>
14 using namespace std;
15 typedef long long LL;
16 const LL inf = 1e18;
17 const int N = 5000;
18 
19 struct Point{
20     int x,y;
21     Point(){}
22     Point(int _x,int _y){
23         x=_x;y=_y;
24     }
25     Point operator -(const Point &b)const{
26         return Point(x-b.x,y-b.y);
27     }
28     int operator *(const Point &b)const{
29         return x*b.x+y*b.y;
30     }
31     int operator ^(const Point &b)const{
32         return x*b.y-y*b.x;
33     } 
34 };
35 
36 struct Line{
37     Point s,e;
38     Line(){}
39     Line(Point _s,Point _e){
40         s=_s,e=_e;
41     }
42 };
43 
44 int xmult(Point p0,Point p1,Point p2){
45     return (p1-p0)^(p2-p0);
46 }
47 
48 Line line[N];
49 int ans[N];
50 int main(){
51     int n,m,x1,y1,x2,y2;
52     bool first = true;
53     while(~scanf("%d",&n)&&n){
54         if(first) first=false;
55         else printf("\n");
56         scanf("%d%d%d%d%d",&m,&x1,&y1,&x2,&y2);
57         for(int i=0;i<n;i++){
58             int x,y;
59             scanf("%d%d",&x,&y);
60             line[i]=Line(Point(x,y1),Point(y,y2));
61         }
62         line[n]=Line(Point(x2,y1),Point(x2,y2));
63         int x,y;
64         Point p;
65         memset(ans,0,sizeof(ans));
66         int tem;
67         while(m--){
68             scanf("%d%d",&x,&y);
69             p = Point(x,y);
70             int l=0,r=n,mid;
71             while(l<=r){
72                 mid=(l+r)>>1;
73                 if(xmult(p,line[mid].s,line[mid].e)<0){
74                     tem=mid;
75                     r=mid-1;
76                 }
77                 else l=mid+1;
78             }
79             ans[tem]++;
80         }
81         for(int i=0;i<=n;i++) printf("%d: %d\n",i,ans[i]);
82     }
83     return 0;
84 }