TOYS - POJ 2318（计算几何，叉积判断）

题目大意：给你一个矩形的左上角和右下角的坐标，然后这个矩形有 N 个隔板分割成 N+1 个区域，下面有 M 组坐标，求出来每个区域包含的坐标数。

 

分析：做的第一道计算几何题目....使用叉积判断方向，然后使用二分查询找到点所在的区域。

 

代码如下：

============================================================================================================================

#include<stdio.h>
#include<math.h>
using namespace std;

const int MAXN = 5e3+7;
const double PI = acos(-1.0);

struct point
{
    double x, y;

    point(int x=0, int y=0):x(x), y(y){}
};
struct Vector
{
    point a, b;

    void InIt(point t1, point t2){a=t1, b=t2;}
    double operator * (const point &p) const
    {
        return (p.x-b.x)*(a.y-b.y) - (p.y-b.y)*(a.x-b.x);
    }
};

Vector line[MAXN];

int Find(int N, point a)
{
    int L=0, R=N;

    while(L <= R)
    {
        int Mid = (L+R) >> 1;

        if(line[Mid] * a < 0)
            R = Mid - 1;
        else
            L = Mid + 1;
    }

    return R;
}

int main()
{
    int M, N;
    double x1, x2, y1, y2, ui, li;

    while(scanf("%d", &N) != EOF && N)
    {
        scanf("%d%lf%lf%lf%lf", &M, &x1, &y1, &x2, &y2);

        int ans[MAXN]={0};

        line[0].InIt(point(x1, y1), point(x1, y2));
        for(int i=1; i<=N; i++)
        {
            scanf("%lf%lf", &ui, &li);
            line[i].InIt(point(ui, y1), point(li, y2));
        }
        while(M--)
        {
            scanf("%lf%lf", &ui, &li);
            int i = Find(N, point(ui, li));

            ans[i] += 1;
        }

        for(int i=0; i<=N; i++)
            printf("%d: %d\n", i, ans[i]);
        printf("\n");
    }

    return 0;
}

 重写...

#include<math.h>
#include<stdio.h>
#include<vector>
#include<iostream>
#include<algorithm>
using namespace std;

const double EPS = 1e-5;
const int maxn = 5005;

int SIGN(const double &val)
{///整数返回1，负数返回-1， 0返回0
    if(val > EPS)return 1;
    if(fabs(val) < EPS)return 0;
    return -1;
}

class Point
{
public:
    Point(double x, double y): x(x), y(y){}
    Point operator- (const Point& other)const
    {///重载减号
        return Point((x-other.x), (y - other.y));
    }
    double operator^(const Point& other)const
    {///重载异或，定义叉积的运算
        return (x*other.y) - (y*other.x);
    }
public:
    double x, y;
};

class Segment
{
public:
    Segment(Point S, Point E) : S(S), E(E){}
    int Mul(Point& other) const
    {///用差乘判断点在线段的方向
        return SIGN( (E-S)^(other-S) );
    }
public:
    Point S, E;
};

class SetSegment
{///定义一个线段的集合，有很多线段构成
public:
    void Insert(const Segment& other)
    {///插入一个线段
        segs.push_back(other);
    }
    unsigned int Find(Point p)
    {///查找点p靠近的最左边的线段的下标
        unsigned int L=0, R=segs.size()-1, M;

        while(L <= R)
        {
            M = (L+R) / 2;
            Segment tmp = segs[M];
            if(tmp.Mul(p) == -1)
                R = M-1;
            else
                L = M+1;
        }

        return R;
    }
public:
    vector<Segment> segs;
};
int main()
{
    int N, M;
    double x1, x2, y1, y2, Ui, Li;

    while(scanf("%d", &N) != EOF && N)
    {
        scanf("%d%lf%lf%lf%lf", &M, &x1, &y1, &x2, &y2);

        SetSegment ss;

        ss.Insert(Segment(Point(x1, y1), Point(x1, y2)));
        for(int i=0; i<N; i++)
        {
            scanf("%lf%lf", &Ui, &Li);
            ss.Insert(Segment(Point(Ui, y1), Point(Li, y2)));
        }

        int ans[maxn] = {0};

        while(M--)
        {
            scanf("%lf%lf", &x1, &y1);

            int index = ss.Find(Point(x1, y1));
            ans[index] += 1;
        }

        for(int i=0; i<=N; i++)
            printf("%d: %d\n", i, ans[i]);
        printf("\n");
    }

    return 0;
}