poj2318——TOYS【计算几何，叉积判断方向】

TOYS Time Limit: 2000MS   Memory Limit: 65536K Total Submissions: 18350   Accepted: 8677 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-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. Source Rocky Mountain 2003

[Submit]   [Go Back]   [Status]   [Discuss]

Home Page   Go Back  To top

---

All Rights Reserved 2003-2013 Ying Fuchen,Xu Pengcheng,Xie Di
Any problem, Please Contact Administrator

这道题的大致意思就是给你一个箱子，把这个箱子分成不同的区域，问每一个区域可以放几个点。

要想解这道题首先要知道一个知识点：如何通过叉积判断点和在直线间的位置。可以参考下面的這篇博客。

https://blog.csdn.net/y990041769/article/details/38258761

然后在每一个区间扫描一遍，以每一个区间的最有边的向量为基准向量，来判断点在这个向量的哪个位置。最后扫描一遍。最后要记住最后一个区间的最后一条边是题目一给出的边，如果所有区间都扫描一遍则落在最后一个区间。

#include
#include
#include
#include
using namespace std;
const int MAXN=10005;
int n,m,x1,y1,x2,y2;
int cnt[MAXN];
struct Point{
    int x,y;
};
struct Line{
    Point a,b;
}line[MAXN];
int det(Point p0,Point p1,Point p2){//p0为向量积的始点
    return (p1.x-p0.x)*(p2.y-p0.y)-(p2.x-p0.x)*(p1.y-p0.y);
}
void judge(Point d){
    for(int i=0;i0) continue;
        else{
            cnt[i]++;
            return ;
        }
    }
    //一直查找到最后一个区域
    cnt[n]++;
    return ;
}
int main(){
    while(~scanf("%d",&n)&&n){
        memset(cnt,0,sizeof(cnt));
        scanf("%d%d%d%d%d",&m,&x1,&y1,&x2,&y2);
        for(int i=0;i