POJ 2318TOYS 叉积的应用

A - TOYS

Time Limit:2000MS     Memory Limit:65536KB     64bit IO Format:%I64d & %I64u

Submit Status Practice POJ 2318

Appoint description: lijunle  (2011-07-18)System Crawler  (2016-05-08)

Description

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-thcardboard 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 03 14 36 810 1015 301 52 12 85 540 107 94 10 0 10 100 020 2040 4060 6080 80 5 1015 1025 1035 1045 1055 1065 1075 1085 1095 100

Sample Output

0: 21: 12: 13: 14: 05: 10: 21: 22: 23: 2
4: 2
给你n条线把一个长方形分割成n+1个空间
给你m个点让你判断这m个点分别在哪些空间中并统计个数
一开始想复杂了 想通过判断点是否在这n+1个4边形中
后来发现只要找到离每个点右边最近的那条线段的编号就可以确定该点属于哪个空间了
这里利用叉积的知识 对于向量 P和Q 
P*Q>0  则P在Q顺时针方向
P*Q<0  则P在Q逆时针方向
P*Q=0  P与Q共线 不一定同向
连接 点 p和 每条线段的下面那个点形成向量就可以通过叉积判断点对于不同线段的位置
这里我们采取二分线段来确定点在的区域
ACcode：
#include <cstring>#include <cstdio>#define maxn 5050struct  Point{    int x,y;    Point (){}    Point(int _x,int _y){        x=_x;        y=_y;    }}p[maxn];struct Line{    Point s,e;    Line(){}    Line(Point _s,Point _e){        s=_s;        e=_e;    }}my[maxn];int ans[maxn];int multi(Point p1,Point  p2,Point p0){    return (p1.x-p0.x)*(p2.y-p0.y)-(p2.x-p0.x)*(p1.y-p0.y);}int main(){    int n,m,x1,y1,x2,y2;    while(~scanf("%d",&n),n){        scanf("%d%d%d%d%d",&m,&x1,&y1,&x2,&y2);        for(int i=0;i<n;++i){            int a,b;            scanf("%d%d",&a,&b);            my[i]=Line(Point(a,y1),Point(b,y2));        }        my[n]=Line(Point(x2,y1),Point(x2,y2));        memset(ans,0,sizeof(ans));        for(int i=1;i<=m;++i){            int x,y;            scanf("%d%d",&x,&y);            Point tmp=Point(x,y);            int l=0,r=n,id;            while(l<=r){                int mid=(l+r)>>1;                if(multi(my[mid].s,my[mid].e,tmp)<0){                    id=mid;                    r=mid-1;                }                else l=mid+1;            }            ans[id]++;        }        for(int i=0;i<=n;++i)            printf("%d: %d\n",i,ans[i]);        putchar('\n');    }    return 0;}