M - Toy Storage解题报告（来自网络）

M - Toy Storage

Time Limit:1000MS     Memory Limit:65536KB     64bit IO Format:%I64d & %I64u

Submit  Status  Practice  POJ 2398

Description

Mom and dad have a problem: their child, Reza, never puts his toys away when he is finished playing with them. They gave Reza a rectangular box to put his toys in. Unfortunately, Reza 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 Reza to find his favorite toys anymore. 
Reza's parents came up with the following idea. They put cardboard partitions into the box. Even if Reza keeps throwing his toys into the box, at least toys that get thrown into different partitions stay separate. The box looks like this from the top: 

We want for each positive integer t, such that there exists a partition with t toys, determine how many partitions have t, toys.

Input

The input consists of a number of cases. The first line consists of six integers n, m, x1, y1, x2, y2. The number of cardboards to form the partitions is n (0 < n <= 1000) and the number of toys is given in m (0 < m <= 1000). 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 each consists of two integers Ui Li, indicating that the ends of the ith cardboard is at the coordinates (Ui, y1) and (Li, y2). You may assume that the cardboards do not intersect with each other. The next m lines each consists of two integers Xi Yi specifying where the ith toy has landed in the box. You may assume that no toy will land on a cardboard. 

A line consisting of a single 0 terminates the input.

Output

For each box, first provide a header stating "Box" on a line of its own. After that, there will be one line of output per count (t > 0) of toys in a partition. The value t will be followed by a colon and a space, followed the number of partitions containing t toys. Output will be sorted in ascending order of t for each box.

Sample Input

4 10 0 10 100 0
20 20
80 80
60 60
40 40
5 10
15 10
95 10
25 10
65 10
75 10
35 10
45 10
55 10
85 10
5 6 0 10 60 0
4 3
15 30
3 1
6 8
10 10
2 1
2 8
1 5
5 5
40 10
7 9
0

Sample Output

Box
2: 5
Box
1: 4
2: 1

题目大意
如图，给你一个长方形box，然后有一些隔板，给你一些点，判断是被装进几号盒子里，

题目分析
叉乘的应用问题，用二分，设在left和right区间内，然后求出mid，若与left的叉乘大于0，与mid的叉乘小于0，则在【left,mid】中，否则在另一半中。

POJ的两道题区别在于
2318所有的隔板都是按顺序给出的，然后需要求出每个隔板区间的有点的个数
2398的，板的顺序是不定的，需要先排序，然后求的是，拥有n个球的盒子个数

这里给出ZOJ 1996的代码

代码示例

#include <iostream>
#include <cstring>
#include <algorithm>
 
using namespace std;
 
struct point
{
	int x,y;
};
 
struct note
{
	point u,d;
}a[5005];
 
void add(int k,int xx,int yy,int xx2,int yy2);
int search(point p,int left,int right);
int cross(point p0,point p1,point p2);
 
bool cmp(note pp,note pp2)
{
	if(pp.u.x==pp2.u.x)
		return pp.d.x<pp2.d.x;
	else
		return pp.u.x<pp2.u.x;
}
 
 
int main()
{
	int m,n,x1,x2,y1,y2,up,down,k,count[5006],cc[5006];
	point p;
	while(cin >> n&&n)
	{
		cin >> m >> x1 >> y1 >> x2 >> y2;
		add(0,x1,y1,x1,y2);
		add(n+1,x2,y2,x2,y1);
		for(int i=1;i<=n;i++)
		{
			cin >> up >> down;
			add(i,up,y1,down,y2);
		}
 
		sort(a+1,a+n+1,cmp);
 
		memset(count,0,sizeof(count));
		memset(cc,0,sizeof(cc));
 
		for(int i=1;i<=m;i++)
		{
			cin >> p.x >> p.y;
			k=search(p,0,n+1);
			count[k]++;
		}
 
		for(int i=0;i<=n;i++)
		{
			if(count[i]!=0)
			cc[count[i]]++;
		}
 
		cout << "Box" << endl;
		for(int i=0;i<=5000;i++)
			if(cc[i]!=0)
				cout << i << ": " << cc[i] << endl; 
 
	}
 
	return 0; 
}
 
void add(int k,int xx,int yy,int xx2,int yy2)
{
	a[k].u.x=xx;
	a[k].u.y=yy;
	a[k].d.x=xx2;
	a[k].d.y=yy2;
}
 
int search(point p,int left,int right)
{
	int mid=(left+right)/2;
	if(left+1>=right)
		return left;
	if(cross(p,a[mid].u,a[mid].d)<0&&(cross(p,a[left].u,a[left].d)>0))
		return search(p,left,mid);
	else
		return search(p,mid,right);
}
 
int cross(point p0,point p1,point p2)
{
	int dx1,dx2,dy1,dy2;
	dx1=p1.x-p0.x;
	dy1=p1.y-p0.y;
	dx2=p2.x-p0.x;
	dy2=p2.y-p0.y;
	return (dx1*dy2-dx2*dy1);
}