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;
}

#include <cstdio>
#include <cstring>
#include <cmath>
#include <iostream>
using namespace std;
#define eps 1e-10
#define N 5005
struct Point{
    double x,y;
    Point(double x=0,double y=0):x(x),y(y){}
};

struct Line{
    Point A , B;
}line[N];
typedef Point Vector;

int dcmp(double x){
    if(abs(x)<eps) return 0;
    else return x<0?-1:1;
}

bool operator == (const Point &a , const Point &b){
    return dcmp(a.x-b.x) == 0 && dcmp(a.y-b.y) == 0;
}

Vector operator+(Vector a , Vector b){
    return Vector(a.x+b.x , a.y+b.y);
}

Vector operator-(Point a, Point b){
    return Vector(a.x-b.x,a.y-b.y);
}

Vector operator*(Vector a,double b){
    return Vector(a.x*b,a.y*b);
}

Vector operator/(Vector a , double b){
    return Vector(a.x/b , a.y /b);
}

double Dot(Vector a,Vector b){
    return a.x * b.x + a.y * b.y;
}

double Cross(Vector a,Vector b){
    return a.x*b.y - a.y*b.x;
}

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;
}

int cnt[N];

int main()
{
    //freopen("test.in","rb",stdin);
    int n,m,x1,y1,x2,y2,ui,li;
    while(scanf("%d",&n)!=EOF){
        if(n == 0) break;

        scanf("%d%d%d%d%d",&m,&x1,&y1,&x2,&y2);

        line[0].A = Point(x1,y2),line[0].B = Point(x1,y1);
        for(int i=1;i<=n;i++){
            scanf("%d%d",&ui,&li);
            line[i].A.x = li , line[i].A.y = y2;
            line[i].B.x = ui , line[i].B.y = y1;
        }
        line[n+1].A = Point(x2,y2) , line[n+1].B = Point(x2,y1);

        memset(cnt,0,sizeof(cnt));

        for(int i=0;i<m;i++){
            scanf("%d%d",&ui,&li);
            Point tmp = Point(ui,li);
            for(int j=0;j<=n;j++){
                if(inArea(tmp,line[j],line[j+1])){
                    cnt[j]++;
                    break;
                }
            }
        }

        for(int i=0;i<=n;i++){
            printf("%d: %d
",i,cnt[i]);
        }
        puts("");
    }
    return 0;
}