文章标题 POJ 2318 ： TOYS （计算几何）

TOYS

Calculate the number of toys that land in each bin of a partitioned toy box.
Mom and dad have a problem - their child John never puts his toys away when he is finished playing with them. They gave John a rectangular box to put his toys in, but John is rebellious and obeys his parents by simply throwing his toys into the box. All the toys get mixed up, and it is impossible for John to find his favorite toys.

John’s parents came up with the following idea. They put cardboard partitions into the box. Even if John keeps throwing his toys into the box, at least toys that get thrown into different bins stay separated. The following diagram shows a top view of an example toy box.

For this problem, you are asked to determine how many toys fall into each partition as John throws them into the toy box.
Input
The input file contains one or more problems. The first line of a problem consists of six integers, n m x1 y1 x2 y2. The number of cardboard partitions is n (0 < n <= 5000) and the number of toys is m (0 < m <= 5000). The coordinates of the upper-left corner and the lower-right corner of the box are (x1,y1) and (x2,y2), respectively. The following n lines contain two integers per line, Ui Li, indicating that the ends of the i-th cardboard partition is at the coordinates (Ui,y1) and (Li,y2). You may assume that the cardboard partitions do not intersect each other and that they are specified in sorted order from left to right. The next m lines contain two integers per line, Xj Yj specifying where the j-th toy has landed in the box. The order of the toy locations is random. You may assume that no toy will land exactly on a cardboard partition or outside the boundary of the box. The input is terminated by a line consisting of a single 0.
Output
The output for each problem will be one line for each separate bin in the toy box. For each bin, print its bin number, followed by a colon and one space, followed by the number of toys thrown into that bin. Bins are numbered from 0 (the leftmost bin) to n (the rightmost bin). Separate the output of different problems by a single blank line.
Sample Input
5 6 0 10 60 0
3 1
4 3
6 8
10 10
15 30
1 5
2 1
2 8
5 5
40 10
7 9
4 10 0 10 100 0
20 20
40 40
60 60
80 80
5 10
15 10
25 10
35 10
45 10
55 10
65 10
75 10
85 10
95 10
0
Sample Output
0: 2
1: 1
2: 1
3: 1
4: 0
5: 1

0: 2
1: 2
2: 2
3: 2
4: 2
Hint
As the example illustrates, toys that fall on the boundary of the box are “in” the box.
题意：在一个平面内有一个平行与x轴矩形，给出左下角和右上角的坐标，然后这个矩形被n条线段被切割成n+1个区域，且线段是从按x坐标升序输入的。默认为任意两条直线不相交，然后有m个点，这m个点落在矩形内，且不会在直线上，问每个区域有多少个点落在里面。
分析：既然线段是按升序输入的，所以可以利用叉积的性质来判断这个点是在线段的左边还是有边，当然第一个区域只需要判断点是否在直线的左边，最后一个区域在第n条直线的有边即可，然后中间的区域就得判断点是否在两条相邻直线之间，即与两条直线的叉积的成积是否小于0。
代码：

#include<iostream>
#include<string>
#include<cstdio>
#include<cstring>
#include<vector>
#include<math.h>
#include<map>
#include<queue> 
#include<algorithm>
using namespace std;
const int inf = 0x3f3f3f3f;
const double eps=1e-8;
int n,m;

//精度函数 
int sgn(double x){
    if (fabs(x)<eps)return 0;
    if (x<0) return -1;
    return 1;
}

struct point {
    double x,y;
    point (double a=0,double b=0){
        x=a;y=b;
    }
    bool operator <(const point &t){
        return (x<t.x)||(x==t.x&&y<t.y);
    }
    //相减 
    point operator - (const point &b)const {
        return point(x-b.x,y-b.y);
    }
    //叉积
    double operator ^ (const point &b)const{
        return x*b.y-y*b.x;
    } 
    //点积
    double operator * (const point &b) const {
        return x*b.x+y*b.y;
    }
}; 

struct line{
    point s,e;
    line(){}
    line (point x,point y){
        s=x,e=y;
    }
    //两直线相交求交点  
    //第一个值为0表示直线重合，为1表示平行，为0表示相交,为2是相交  
    //只有第一个值为2时，交点才有意义
    pair<int,point> operator &(const line &b)const{         
        point res = s;         
        if(sgn((s-e)^(b.s-b.e)) == 0){             
            if(sgn((s-b.e)^(b.s-b.e)) == 0)return make_pair(0,res);//重合             
            else return make_pair(1,res);//平行         
        }
        double t = ((s-b.s)^(b.s-b.e))/((s-e)^(b.s-b.e));
        res.x += (e.x-s.x)*t;
        res.y += (e.y-s.y)*t;         
        return make_pair(2,res);     
    } 
};
double x1;
double y,x2,y2;
double px1[5005],px2[5005];//每条线段的两个x坐标 
point P[5005];//每个点的位置 
int num[5005];//每个区域的数目 
int vis[5005];//标记，判断这个点是否已经属于哪个区域里了 

int main ()
{
    while (scanf ("%d",&n)&&n){
        scanf ("%d",&m);
        scanf ("%lf%lf%lf%lf",&x1,&y,&x2,&y2);
        for (int i=1;i<=n;i++){
            scanf ("%lf%lf",&px1[i],&px2[i]);
        } 
        for (int i=1;i<=m;i++){
            scanf ("%lf%lf",&P[i].x,&P[i].y);
        }
        memset (vis,0,sizeof (vis));
        memset (num,0,sizeof (num));
        for (int i=1;i<=m;i++){
            //判断在第0个区域，只需判断是否在第一条边的左边就行 
            if (sgn(((P[i]-point(px2[1],y2))^(point(px1[1],y)-point(px2[1],y2))))<0){
                num[0]++;
                vis[i]=1;
            }
        }
        for (int i=1;i<=n-1;i++){
            for (int j=1;j<=m;j++){
                if (vis[j])continue;
                //接下来的区域得判断在两条线段之间，即叉积结果的成积小于0 
                double t1=sgn((P[j]-point(px2[i],y2))^(point(px1[i],y)-point(px2[i],y2)));
                double t2=sgn((P[j]-point(px2[i+1],y2))^(point(px1[i+1],y)-point(px2[i+1],y2)));
                if (t1*t2<0){
                    num[i]++;
                    vis[j]=1;
                }
            }
        }
        for (int i=1;i<=m;i++){
            if (vis[i])continue;
            //最后一个区域只需要判断是否在第n条线段的有边即可 
            if (sgn(((P[i]-point(px2[n],y2))^(point(px1[n],y)-point(px2[n],y2))))>0){
                num[n]++;
                vis[i]=1;
            }
        }
        for (int i=0;i<=n;i++){
            printf ("%d: %d\n",i,num[i]);
        }
        printf ("\n");
    }
    return 0;
}