POJ 2318 TOYS（计算几何）

A - TOYS

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

Submit  Status

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.

题目大意：有一个矩形盒子，盒子里会有一些木块线段，并且这些线段是按照顺序给出的，有n条线段，把盒子分层了n+1个区域,然后有m个玩具，这m个玩具的坐标是已知的，问最后每个区域有多少个玩具

解题思路：因为线段是有序给出，所以不用排序，判断某个点在哪个区域，采用二分法，将某个点和线段的叉积来判断这个点是在线的左边或者右边，根据这个来二分找出区域

代码和思路参考与此链接：http://blog.youkuaiyun.com/wangjian8006

这也算我入门计算几何的第一道题了，首先看了一些有关资料，了解到叉积的重要性！！有2点，p1，p2构成一条线段L，L左边有一点p3，右边有一点p4，则（p1-p2）x（p3-p1）>0，（p1-p2）x（p4-p1）<0。

也就是说在L左边的点叉积>0，右边的<0。

剩下的就是二分找答案了，在记录到数组中，最后输出。

#include <iostream>
using namespace std;
#define _MAXL_ 5010

typedef struct{
	int x, y;
}POINT;

typedef struct{
	POINT first,second;
}LINE;

class CBox{
private:
	int nTop,nLeft,nRight,nButtom;
	int nCountLine;						//记录盒子里有多少条线
	LINE *pLine;						//记录线的信息
	int *pRange;						//记录区间中的玩具有多少个
	int Multi(POINT p1,POINT p2,POINT p0);			//叉积计算
public:
	CBox();
	virtual ~CBox();
	void SetBorder(int top,int left,int right,int buttom);		//设置盒子矩形坐标
	void AddLine(int first,int second);							//在盒子中增加一条线
	int GetRange(int val);								//得到某个区间中的玩具个数
	void SetToy(POINT toy);							//在盒子里放一个玩具
};

CBox::CBox(){
	nCountLine = 0;
	pLine = new LINE[_MAXL_];
	pRange = new int[_MAXL_];
	memset(pRange,0,sizeof(pRange)*_MAXL_);
}

CBox::~CBox(){
	delete []pLine;
	delete []pRange;
}

void CBox::SetBorder(int top,int left,int right,int buttom){
	nTop = top;
	nLeft = left;
	nRight = right;
	nButtom = buttom;
}

void CBox::AddLine(int topx,int buttomx){
	pLine[nCountLine].first.x = topx;
	pLine[nCountLine].first.y = nTop;
	pLine[nCountLine].second.x = buttomx;
	pLine[nCountLine].second.y = nButtom;
	nCountLine++;
}

int CBox::GetRange(int val){
	return pRange[val];
}

int CBox::Multi(POINT p1,POINT p2,POINT p0){
	 return (p1.x - p0.x) * (p2.y - p0.y) - (p2.x - p0.x) * (p1.y - p0.y);
}

void CBox::SetToy(POINT toy){			//利用二分查找，查找出该点在哪个区间
	int l, r, mid;
	l=0,r=nCountLine-1;
	while (l < r){
		mid = (l+r)>>1;
		if (Multi(toy, pLine[mid].first, pLine[mid].second) > 0) l = mid + 1;
		else r = mid;
	}  
	if (Multi(toy, pLine[l].first, pLine[l].second) < 0) pRange[l]++;
	else pRange[l+1]++;
}

int main(){
	int i,fx,lx;
	POINT a,b;
	int n,m;
	while(cin>>n && n){
		cin>>m>>a.x>>a.y>>b.x>>b.y;
		CBox box;
		box.SetBorder(a.y,a.x,b.x,b.y);
		for(i = 0;i < n;i++){
			cin>>fx>>lx;
			box.AddLine(fx,lx);
		}

		for(i = 0;i < m;i++){
			cin>>a.x>>a.y;
			box.SetToy(a);
		}

		for(i = 0;i <= n;i++){
			cout<<i<<": "<<box.GetRange(i)<<endl;
		}
		cout<<endl;
	}
	return 0;
}