POJ 2318 TOYS (计算几何)

题目

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, $n$ $m$ $x_1$ $y_1$ $x_2$ $y_2$ . The number of cardboard partitions is $n$ $(0 < n \leq 5000)$ and the number of toys is $m$ $(0 < m \leq 5000)$ . The coordinates of the upper-left corner and the lower-right corner of the box are $(x_1,y_1)$ and $(x_2,y_2)$ , respectively. The following $n$ lines contain two integers per line, $U_i$ $L_i$ , indicating that the ends of the i-th cardboard partition is at the coordinates $(U_i,y_1)$ and $(L_i,y_2)$ . 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, $X_j$ $Y_j$ 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

有一个玩具盒,被 $n$ 个隔板分开成左到 $u$ 右 $n+1$ 个区域,给出每个左上角和右下角边界点的坐标,给出每个分割线两个端点的坐标,然后给每个玩具的坐标,求每个区域有几个玩具。


分析

判断点在直线的左右利用向量叉乘就可以判断,从线段的一个起点出发到判断点构成的向量和从判断点到另一个端点组成的向量进行叉乘,如果叉乘大于 $0$ ,那么点在直线(线段)右边,否则在直线(线段)左边。


知识

设 $\vec a=(x_1,y_1)$ , $\vec b=(x_2,y_2)$

点乘:内积(数量积),$\vec{a}\cdot\vec{b}=|a||b|cos\theta=x_1\cdot x_2+y_1\cdot y_2$

叉乘:外积(向量积),$|\vec{c}|=|\vec{a}\times\vec{b}|=|a||b|sin\theta=x_1\cdot y_2-x_2\cdot y_1$

向量积的结果是一个向量,方向用“右手法则”判断(四指为a的方向,朝手心方向摆动到b的方向,大拇指就是c的方向)

几何意义:

点乘的几何意义是:是一条边向另一条边的投影乘以另一条边的长度

叉乘的几何意义是:两个矢量围成的平行四边形的面积


代码

#include<iostream>
#include<cstdio>
using namespace std;
typedef long long ll;
#define INF 0x3f3f3f3f
#define mod 1e9+7
#define clr(a,x) memset(a,x,sizeof(a))
const double eps = 1e-6;
const int N = 50001;
int n,m;
int x1,y1,x2,y2,u[N],l[N];
int ans[N];
int x,y;
int xmul(int x1,int y1,int x2,int y2)
{
    return (x1*y2)-(x2*y1);
}
int main()
{
    while(cin>>n,n)
    {
        cin>>m>>x1>>y1>>x2>>y2;
        for(int i=0;i<n;i++)
        {
            cin>>u[i]>>l[i];
            ans[i]=0;
        }
        ans[n]=0;
        for(int i=0,j;i<m;i++)  
        {
            cin>>x>>y;
            for(j=0;j<n;j++)
                if(xmul(x-l[j],y-y2,u[j]-l[j],y1-y2)<=0)
                    break;
            ans[j]++;
        }
        for(int i=0;i<=n;i++)
        {
            printf("%d: %d\n",i,ans[i]);
        }
        cout<<endl;
    }
    return 0;
}
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Last modification:November 2nd, 2019 at 02:35 pm
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