POJ 1755 Triathlon 半平面交 - c++编程基础

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   
#include   
#include   
#include   
#include   
#include   
#include   
#include   
#include   
#include   
#include   
#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;  
}