POJ 2398 Toy Storage(计算几何)

标签：

题意：给定一个如上的长方形箱子，中间有n条线段，将其分为n+1个区域，给定m个玩具的坐标，统计每个区域中的玩具个数。

题解：通过斜率判断一个点是否在两条线段之间。

/**
通过斜率比较点是否在两线段之间
*/
 
#include"iostream"
#include"cstdio"
#include"algorithm"
#include"cstring"
using namespace std;
const int N=1005;  
 
struct edgeP                                      //边上的一个点
{
    int x1,x2;
}e[N];
 
struct point
{
    int x,y;
}p[N];
 
int cmp(edgeP a,edgeP b)
{
    return a.x1<=b.x1;
}
 
int x1,y1,x2,y2;
 
bool is_z(point e1,point e2)                // /型斜线
{
    if((e1.y-e2.y)*(e1.x-e2.x)>=0)
        return true;
    else
        return false;
}
 
bool is_f(point e1,point e2)                  // \型斜线
{
    if((e1.y-e2.y)*(e1.x-e2.x)<=0)
        return true;
    else
        return false;
}
 
bool is_inzr(point e1,point e2,point p)            //   在/型斜线的右边
{
    if((e1.y-e2.y)*(e1.x-e2.x)>=0)
    {
        if((p.x-e2.x>0)&&(e1.y-e2.y)*(p.x-e2.x)>(p.y-e2.y)*(e1.x-e2.x))
            return true;
    }
    return false;
}
 
bool is_infl(point e1,point e2,point p)           //   在\型斜线的左边
{
    if((e1.y-e2.y)*(e1.x-e2.x)<=0)
    {
        if((p.x-e2.x<0)&&(e1.y-e2.y)*(p.x-e2.x)<(p.y-e2.y)*(e1.x-e2.x))
            return true;
    }
    return false;
}
 
bool is_in(point e1,point e2,point e3,point e4,point p)             // 点是否在两线内
{
    if((is_z(e1,e2)&&is_inzr(e1,e2,p))&&(is_f(e3,e4)&&is_infl(e3,e4,p)))   //  点在/.\型两线间
        return true;
    if((is_z(e1,e2)&&is_inzr(e1,e2,p))&&(is_z(e3,e4)&&!is_inzr(e3,e4,p)))   //  点在/./型两线间
        return true;
    if((is_f(e1,e2)&&!is_infl(e1,e2,p))&&(is_f(e3,e4)&&is_infl(e3,e4,p)))   //点在\.\型两线间
        return true;
    if((is_f(e1,e2)&&!is_infl(e1,e2,p))&&(is_z(e3,e4)&&!is_inzr(e3,e4,p)))   //点在\./型两线间
        return true;
    return false;
}
 
int main()
{
    int n,m;
    while(cin>>n)
    {
        if(n==0)
            return 0;
        e[0].x1=0,e[0].x2=0;
        cin>>m>>x1>>y1>>x2>>y2;
        for(int i=1;i<=n;i++)
        {
            scanf("%d%d",&e[i].x1,&e[i].x2);
        }
        for(int i=0;i<m;i++)
        {
            scanf("%d%d",&p[i].x,&p[i].y);
        }
        e[n+1].x1=x2,e[n+1].x2=x2;
        sort(e,e+n+2,cmp);
        int cnt[N];
        memset(cnt,0,sizeof(cnt));
        for(int i=0;i<=n;i++)
        {
            for(int j=0;j<m;j++)
            {
 
                point e1,e2;
                e1.x=e[i].x1,e1.y=y1;
                e2.x=e[i].x2,e2.y=y2;
                point e3,e4;
                e3.x=e[i+1].x1,e3.y=y1;
                e4.x=e[i+1].x2,e4.y=y2;
                /*{
                    cout<<'('<<e1.x<<','<<e1.y<<')'<<"   "<<'('<<e2.x<<','<<e2.y<<')'<<endl;
                    cout<<'('<<e3.x<<','<<e3.y<<')'<<"   "<<'('<<e4.x<<','<<e4.y<<')'<<endl;
                    cout<<'('<<p[j].x<<','<<p[j].y<<')'<<endl;
                }*/
                if(is_in(e1,e2,e3,e4,p[j]))
                {
                    cnt[i]++;
                    //cout<<"cnt"<<i<<"++++++++++++++++++++++"<<endl;
                }
            }
            //cout<<"-------------------------------------------"<<endl;
        }
        /*for(int i=0;i<=n;i++)
        {
            cout<<cnt[i]<<' ';
        }*/
        sort(cnt,cnt+n+1);
        puts("Box");
        int j=cnt[0],count=1;
        cnt[n+1]=-10;
        for(int i=1;i<=n+1;i++)
        {
            if(cnt[i]==j)
            {
                count++;
            }
            else
            {
                if(j!=0)
                    printf("%d: %d\n",j,count);
                j=cnt[i];
                count=1;
            }
        }
    }
}

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

POJ 2398 Toy Storage(计算几何)

标签：

原文地址：http://blog.csdn.net/strangedbly/article/details/51363836