POJ 2318 TOYS （计算几何）叉积的运用

Posted 2020-10-12 华东交通ACM训练基地

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.

题意：已知一个左上角顶点（x1,y1），右下角顶点（x2,y2）的方形区域，区域被n条线分成n+1个区域，题目保证线从左到右给出且不相交。

告诉你m个点，点不会在线上出现，求各个区域内的点的数量。

叉积的初步运用，数据也很松，这道题目只需要将点与线依次判断叉积即可。

如何判断点s（x0,y0）在s1（x1,y1）,s2（x2,y2）的那一侧？

先要得到向量p1=s1-s0=（x1-x0,y1-y0），p2=s2-s0=（x2-x0,y2-y0），这里再用叉积的公式即可，结果依据右手定则。

#include<iostream>
#include<cstdio>
#include<cstring>
#include<algorithm>
#include<cmath>
#include<stack>
#include<map>
#include<vector>
#include<set>
using namespace std;
const int MAX=5e3+10;
const double eps=1e-4;
const double mod=1e9+7;
#define INF 0x7fffffff
#define ll long long
#define edl putchar(‘\n‘)
#define useit  ios::sync_with_stdio(false);cin.tie(0);cout.tie(0);
#define FOR(i,a,b) for(int i=a;i<=b;i++)
#define ROF(i,a,b) for(int i=a;i>=b;i--)
#define mst(a) memset(a,0,sizeof(a))
#define mstn(a,n) memset(a,n,sizeof(a))
struct point{double x,y;}a[MAX][3],b;
//bool cmp(const num &x, const num &y){return x.v>y.v;}
double cross(point a,point b){return (a.x*b.y-a.y*b.x);}
double dot(point a,point b){return (a.x*b.x+a.y*b.y);}
double cro(point a,point b,point c){return (a.x-c.x)*(b.y-c.y)-(b.x-c.x)*(a.y-c.y);}
int main()
{
    double x1,y1,x2,y2;
    int n,m,cas=0;
    int c[MAX];
    while(scanf("%d",&n)&&n)
    {
        scanf("%d%lf%lf%lf%lf",&m,&x1,&y1,&x2,&y2);
        if(cas) edl;
        mst(c);
        FOR(i,1,n)cin>>a[i][1].x>>a[i][2].x,a[i][1].y=y1,a[i][2].y=y2;
        //a[0][1].x=x1,a[0][1].y=y1,a[0][2].x=x1,a[0][2].y=y2;
        a[n+1][1].x=x2,a[n+1][1].y=y1,a[n+1][2].x=x2,a[n+1][2].y=y2;
        while(m--)
        {
            cin>>b.x>>b.y;
            FOR(i,1,n+1)
            {
                if(cro(a[i][2],a[i][1],b)>0)
                {
                    c[i-1]++;
                    break;
                }
            }
        }
        FOR(i,0,n)
        cout<<i<<": "<<c[i]<<endl;
        cas++;
    }
    
}

 至于第二题，题面几乎完全相同，不同点有二：一、线的给出不按从左到右，但仍不相交。二、答案要求略有不同

解决线的排序问题很简单，由于不会相交，只要按线的中点坐标排序即可。

代码如下：

#include<iostream>
#include<cstdio>
#include<cstring>
#include<algorithm>
#include<cmath>
#include<stack>
#include<map>
#include<vector>
#include<set>
using namespace std;
const int MAX=5e3+10;
const double eps=1e-4;
const double mod=1e9+7;
#define INF 0x7fffffff
#define ll long long
#define edl putchar(‘\n‘)
#define useit  ios::sync_with_stdio(false);cin.tie(0);cout.tie(0);
#define FOR(i,a,b) for(int i=a;i<=b;i++)
#define ROF(i,a,b) for(int i=a;i>=b;i--)
#define mst(a) memset(a,0,sizeof(a))
#define mstn(a,n) memset(a,n,sizeof(a))
struct point{double x1,y1,x2,y2,c;}a[MAX],b;
bool cmp(const point &x, const point &y){return x.c<y.c;}
//double cross(point a,point b){return (a.x*b.y-a.y*b.x);}
//double dot(point a,point b){return (a.x*b.x+a.y*b.y);}
double cro(point a,point c){return (a.x2-c.x1)*(a.y1-c.y1)-(a.x1-c.x1)*(a.y2-c.y1);}
int main()
{
    double x1,y1,x2,y2;
    int n,m,cas=0;
    int c[MAX],d[MAX],e[MAX];
    while(scanf("%d",&n)&&n)
    {
        scanf("%d%lf%lf%lf%lf",&m,&x1,&y1,&x2,&y2);
        if(cas) edl;
        mst(c);
        mst(d);
        FOR(i,1,n)cin>>a[i].x1>>a[i].x2,a[i].y1=y1,a[i].y2=y2,a[i].c=a[i].x1+a[i].x2;
        sort(a+1,a+n+1,cmp);
        //a[0][1].x=x1,a[0][1].y=y1,a[0][2].x=x1,a[0][2].y=y2;
        a[n+1].x1=x2,a[n+1].y1=y1,a[n+1].x2=x2,a[n+1].y2=y2;
        while(m--)
        {
            cin>>b.x1>>b.y1;
            FOR(i,1,n+1)
            {
                if(cro(a[i],b)>0)
                {
                    c[i-1]++;
                    break;
                }
            }
        }
        sort(c,c+n+1);
        printf("Box\n");
        int jishu=1;
        FOR(i,0,n)
        {
            if(c[i]!=0)
            {
                if(c[i]==c[i+1])
                jishu++;
                else
                {
                    printf("%d: %d\n",c[i],jishu);
                    jishu=1;
                }
            }
        }
        cas++;
    }        
}