POJ 2318 TOYS(几何)

最新推荐文章于 2020-08-17 23:33:52 发布

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本文深入探讨了使用点积判断点相对于直线位置的方法，通过二分查找优化搜索效率，适用于排序后的点集。详细解释了算法实现过程及其实例应用。

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题目链接：POJ 2318 TOYS

用点积判断一个点在直线的哪一侧，由于题目给出的分割线是排序后的，那么可以直接用二分得出答案。

#include <iostream>
#include <cstdio>
#include <cmath>
#include <cstring>

using namespace std;

const double eps = 1e-10;
const int MAX_N = 5000 + 10;

struct Point
{
	double x, y;
	Point(double x=0, double y=0):x(x),y(y) { }
};

typedef Point Vector;

Vector operator + (const Vector& A, const Vector& B)
{
    return Vector(A.x+B.x, A.y+B.y);
}

Vector operator - (const Point& A, const Point& B)
{
    return Vector(A.x-B.x, A.y-B.y);
}

Vector operator * (const Vector& A, double p)
{
    return Vector(A.x*p, A.y*p);
}

bool operator < (const Point& a, const Point& b)
{
	return a.x < b.x || (a.x == b.x && a.y < b.y);
}

int dcmp(double x)
{
    if(fabs(x) < eps)
        return 0;
    else
        return x < 0 ? -1 : 1;
}

bool operator == (const Point& a, const Point &b)
{
	return dcmp(a.x-b.x) == 0 && dcmp(a.y-b.y) == 0;
}

double Cross(const Vector& A, const Vector& B)
{
    return A.x*B.y - A.y*B.x;
}

Point p[MAX_N << 1];
Point bound[3];
Point toy;

int _count[MAX_N];
int n, m;

void b_search()
{
    int l = 0, r = n - 1, mid;
    while(l <= r)
    {
        mid = (l + r) >> 1;
        if(Cross(toy - p[mid], p[mid + n] - p[mid]) > 0)
            r = mid - 1;
        else if(Cross(toy - p[mid], p[mid + n] - p[mid]) < 0)
            l = mid + 1;
    }

    _count[r + 1]++;
}

int main()
{
    //freopen("in.txt", "r", stdin);
    while(scanf("%d", &n), n)
    {
        memset(_count, 0, sizeof(_count));
        scanf("%d", &m);
        scanf("%lf%lf%lf%lf", &bound[0].x, &bound[0].y, &bound[2].x, &bound[2].y);
        for(int i = 0; i < n; i++)
        {
            scanf("%lf%lf",&p[i].x, &p[i + n].x);
            p[i].y = bound[0].y;
            p[i + n].y = bound[2].y;
        }
        for(int i = 0; i < m; i++)
        {
            scanf("%lf%lf", &toy.x, &toy.y);
            b_search();
        }
        for(int i = 0; i <= n; i++)
            printf("%d: %d\n", i, _count[i]);
        printf("\n");
    }
    return 0;
}