POJ 1755 Triathlon 半平面交

看的这里：http://blog.csdn.net/non_cease/article/details/7820361

题意：铁人三项比赛，给出ｎ个人进行每一项的速度vi, ui, wi;  对每个人判断，通过改变3项比赛的路程，是否能让该人获胜（严格获胜）。

思路：题目实际上是给出了n个式子方程，Ti  = Ai * x + Bi * y + Ci * z ， 0 < i < n

          要判断第i个人能否获胜，即判断不等式组   Tj - Ti > 0,      0 < j < n && j != i    有解

        即 （Aj - Ai）* x + （Bj - Bi） * y + ( Cj - Ci ) * z > 0,   0 < j < n && j != i 有解

         由于 z > 0, 所以 可以两边同时除以 z， 将 x / z, y / z 分别看成 x和 y ， 这样就化三维为二维，可用半平面交判断是否存在解了，

         对每个人构造一次，求一次半平面交即可。

关键是根据这个斜率式子怎么搞成向量的。需要想一想。

然后注意的是半平面交出来是单独一个点是不行的。

因为题目要求的是严格胜出

 

#include <iostream>
#include <cstdio>
#include <cstring>
#include <string>
#include <algorithm>
#include <cstdlib>
#include <cmath>
#include <map>
#include <sstream>
#include <queue>
#include <vector>
#define MAXN 111111
#define MAXM 211111
#define PI acos(-1.0)
#define eps 1e-8
#define INF 1e10
using namespace std;
int dblcmp(double d)
{
    if (fabs(d) < eps) return 0;
    return d > eps ? 1 : -1;
}
struct point
{
    double x, y;
    point(){}
    point(double _x, double _y):
    x(_x), y(_y){};
    void input()
    {
        scanf("%lf%lf",&x, &y);
    }
    double dot(point p)
    {
        return x * p.x + y * p.y;
    }
    double distance(point p)
    {
        return hypot(x - p.x, y - p.y);
    }
    point sub(point p)
    {
        return point(x - p.x, y - p.y);
    }
    double det(point p)
    {
        return x * p.y - y * p.x;
    }
    bool operator == (point a)const
    {
        return dblcmp(a.x - x) == 0 && dblcmp(a.y - y) == 0;
    }
    bool operator < (point a)const
    {
        return dblcmp(a.x - x) == 0 ? dblcmp(y - a.y) < 0 : x < a.x;
    }

}p[MAXN];
struct line
{
    point a,b;
    line(){}
    line(point _a,point _b)
    {
        a=_a;
        b=_b;
    }
    bool parallel(line v)
    {
        return dblcmp(b.sub(a).det(v.b.sub(v.a))) == 0;
    }
    point crosspoint(line v)
    {
        double a1 = v.b.sub(v.a).det(a.sub(v.a));
        double a2 = v.b.sub(v.a).det(b.sub(v.a));
        return point((a.x * a2 - b.x * a1) / (a2 - a1), (a.y * a2 - b.y * a1) / (a2 - a1));
    }
    bool operator == (line v)const
    {
    	return (a == v.a) && (b == v.b);
    }
};
struct halfplane:public line
{
	double angle;
	halfplane(){}
	//表示向量 a->b逆时针(左侧)的半平面
	halfplane(point _a, point _b)
	{
		a = _a;
		b = _b;
	}
	halfplane(line v)
	{
		a = v.a;
		b = v.b;
	}
	void calcangle()
	{
		angle = atan2(b.y - a.y, b.x - a.x);
	}
	bool operator <(const halfplane &b)const
	{
		return dblcmp(angle - b.angle) < 0;
	}
};
struct polygon
{
    int n;
    point p[MAXN];
    line l[MAXN];
    double area;
    void getline()
    {
        for (int i = 0; i < n; i++)
        {
            l[i] = line(p[i], p[(i + 1) % n]);
        }
    }
    void getarea()
    {
        area = 0;
        int a = 1, b = 2;
        while(b <= n - 1)
        {
            area += p[a].sub(p[0]).det(p[b].sub(p[0]));
            a++;
            b++;
        }
        area = fabs(area) / 2;
    }
}convex;
bool judge(point a, point b, point o)
{
    return dblcmp(a.sub(o).det(b.sub(o))) <= 0; //此处有等于号代表的是求出的半平面交为一个点不合法，去掉等于号则代表交成一个点也行
}
struct halfplanes
{
	int n;
	halfplane hp[MAXN];
	point p[MAXN];
	int que[MAXN];
	int st, ed;
	void push(halfplane tmp)
	{
		hp[n++] = tmp;
	}
	void unique()
	{
		int m = 1, i;
		for (i = 1; i < n;i++)
		{
			if (dblcmp(hp[i].angle - hp[i - 1].angle))hp[m++] = hp[i];
			else if (dblcmp(hp[m - 1].b.sub(hp[m - 1].a).det(hp[i].a.sub(hp[m - 1].a)) > 0))hp[m - 1] = hp[i];
		}
		n = m;
	}
	bool halfplaneinsert()
	{
		int i;
		for (i = 0; i < n; i++) hp[i].calcangle();
		sort(hp, hp + n);
		unique();
		que[st = 0] = 0;
		que[ed = 1] = 1;
		p[1] = hp[0].crosspoint(hp[1]);
		for (i = 2; i < n; i++)
		{
			while (st < ed && judge(hp[i].b, p[ed], hp[i].a)) ed--;
			while (st < ed && judge(hp[i].b, p[st + 1], hp[i].a)) st++;
			que[++ed] = i;
			if (hp[i].parallel(hp[que[ed - 1]])) return false;
			p[ed] = hp[i].crosspoint(hp[que[ed - 1]]);
		}
		while (st < ed && judge(hp[que[st]].b, p[ed], hp[que[st]].a)) ed--;
		while (st < ed && judge(hp[que[ed]].b, p[st + 1], hp[que[ed]].a)) st++;
		if (st + 1 >= ed)return false;
		return true;
	}
	void getconvex(polygon &con)
	{
		p[st] = hp[que[st]].crosspoint(hp[que[ed]]);
		con.n = ed - st + 1;
		int j = st, i = 0;
		for (; j <= ed; i++, j++)
		{
			con.p[i] = p[j];
		}
	}
}h;
int A[MAXN], B[MAXN], C[MAXN];
int n;
int main()
{
    double xa, xb, ya, yb;
    scanf("%d", &n);
    for(int i = 0; i < n; i++) scanf("%d%d%d", &A[i], &B[i], &C[i]);

    for(int i = 0; i < n; i++)
    {
        int flag = 0;
        h.n = 0;
        h.push(halfplane(point(0, 0), point(INF, 0)));
        h.push(halfplane(point(INF, 0), point(INF, INF)));
        h.push(halfplane(point(INF, INF), point(0, INF)));
        h.push(halfplane(point(0, INF), point(0, 0)));

        for(int j = 0; j < n; j++)
        {
            if(j == i) continue;
            double a = 1.0 / A[j] - 1.0 / A[i];
            double b = 1.0 / B[j] - 1.0 / B[i];
            double c = 1.0 / C[j] - 1.0 / C[i];
            int d1 = dblcmp(a);
            int d2 = dblcmp(b);
            int d3 = dblcmp(c);
            if(!d1)
            {
                if(!d2)
                {
                    if(d3 <= 0)
                    {
                        flag = 1;
                        break;
                    }
                    continue;
                }
                xa = 0, xb = d2;
                ya = yb = -c / b;
            }
            else
            {
                if(!d2)
                {
                    xa = xb = -c / a;
                    ya = 0, yb = -d1;
                }
                else
                {
                    xa = 0;
                    ya = -c / b;
                    xb = d2;
                    yb = -(c + a * xb) / b;
                }
            }
            h.push(halfplane(point(xa, ya), point(xb, yb)));
        }
        if(flag || !h.halfplaneinsert() ) puts("No");
        else puts("Yes");
    }
    return 0;
}