POJ 2318

题目大意：
箱子有一堆木板隔离开不同区域

给定木板的起末点位置，和一堆物品入箱的坐标，最后来求每块区域的物品个数

 

这里我们可以很容易得知，一个物品的点所在的区域，和前后两个木板形成的叉积值正负性是正好相反的，所以函数如下：

bool inArea(Point a , Line L1 , Line L2){
    double t1 = Cross(a-L1.A , L1.B - L1.A) , t2 = Cross(a-L2.A , L2.B - L2.A);
    //cout<<"Cross: "<<t1<<" "<<t2<<endl;
    return dcmp(t1)*dcmp(t2) < 0;
}

 

 1 #include <cstdio>
 2 #include <cstring>
 3 #include <cmath>
 4 #include <iostream>
 5 using namespace std;
 6 #define eps 1e-10
 7 #define N 5005
 8 struct Point{
 9     double x,y;
10     Point(double x=0,double y=0):x(x),y(y){}
11 };
12 
13 struct Line{
14     Point A , B;
15 }line[N];
16 typedef Point Vector;
17 
18 int dcmp(double x){
19     if(abs(x)<eps) return 0;
20     else return x<0?-1:1;
21 }
22 
23 bool operator == (const Point &a , const Point &b){
24     return dcmp(a.x-b.x) == 0 && dcmp(a.y-b.y) == 0;
25 }
26 
27 Vector operator+(Vector a , Vector b){
28     return Vector(a.x+b.x , a.y+b.y);
29 }
30 
31 Vector operator-(Point a, Point b){
32     return Vector(a.x-b.x,a.y-b.y);
33 }
34 
35 Vector operator*(Vector a,double b){
36     return Vector(a.x*b,a.y*b);
37 }
38 
39 Vector operator/(Vector a , double b){
40     return Vector(a.x/b , a.y /b);
41 }
42 
43 double Dot(Vector a,Vector b){
44     return a.x * b.x + a.y * b.y;
45 }
46 
47 double Cross(Vector a,Vector b){
48     return a.x*b.y - a.y*b.x;
49 }
50 
51 bool inArea(Point a , Line L1 , Line L2){
52     double t1 = Cross(a-L1.A , L1.B - L1.A) , t2 = Cross(a-L2.A , L2.B - L2.A);
53     //cout<<"Cross: "<<t1<<" "<<t2<<endl;
54     return dcmp(t1)*dcmp(t2) < 0;
55 }
56 
57 int cnt[N];
58 
59 int main()
60 {
61     //freopen("test.in","rb",stdin);
62     int n,m,x1,y1,x2,y2,ui,li;
63     while(scanf("%d",&n)!=EOF){
64         if(n == 0) break;
65 
66         scanf("%d%d%d%d%d",&m,&x1,&y1,&x2,&y2);
67 
68         line[0].A = Point(x1,y2),line[0].B = Point(x1,y1);
69         for(int i=1;i<=n;i++){
70             scanf("%d%d",&ui,&li);
71             line[i].A.x = li , line[i].A.y = y2;
72             line[i].B.x = ui , line[i].B.y = y1;
73         }
74         line[n+1].A = Point(x2,y2) , line[n+1].B = Point(x2,y1);
75 
76         memset(cnt,0,sizeof(cnt));
77 
78         for(int i=0;i<m;i++){
79             scanf("%d%d",&ui,&li);
80             Point tmp = Point(ui,li);
81             for(int j=0;j<=n;j++){
82                 if(inArea(tmp,line[j],line[j+1])){
83                     cnt[j]++;
84                     break;
85                 }
86             }
87         }
88 
89         for(int i=0;i<=n;i++){
90             printf("%d: %d\n",i,cnt[i]);
91         }
92         puts("");
93     }
94     return 0;
95 }

 

 posted on 2014-10-10 09:51  Love风吟  阅读(201)  评论(0)  编辑  收藏  举报