POJ 1755 Triathlon （半平面交应用 + 思维 + 直线用两点的向量表示）

#include <iostream>
#include <stdio.h>
#include <string.h>
#include <algorithm>
#include <queue>
#include <map>
#include <vector>
#include <set>
#include <string>
#include <math.h>

#define LL long long
#define mem(i, j) memset(i, j, sizeof(i))
#define rep(i, j, k) for(int i = j; i <= k; i++)
#define dep(i, j, k) for(int i = k; i >= j; i--)
#define pb push_back
#define make make_pair
#define INF INT_MAX
#define inf LLONG_MAX
#define PI acos(-1)
#define fir first
#define sec second
using namespace std;

const int N = 150;
const double eps = 1e-10;
const double maxL = 1e10;

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

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

Point operator + (Point A, Point B) { return Point(A.x + B.x, A.y + B.y); }
Point operator - (Point A, Point B) { return Point(A.x - B.x, A.y - B.y); }
Point operator * (Point A, double p) { return Point(A.x * p, A.y * p); }
Point operator / (Point A, double p) { return Point(A.x / p, A.y / p); }

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 Length(Point A) { return sqrt(Dot(A, A)); }
/* 有向直线，它的左边就是对应的半平面 */
struct Line {
    Point p; /// 直线任意一点
    Point v; /// 方向向量
    double ang; /// 极角，即从x正半轴旋转到向量v所需要的角（弧度）
    Line() { }
    Line(Point p, Point v) : p(p), v(v) { ang = atan2(v.y, v.x); }
    bool operator < (const Line& L) const {
        return ang < L.ang;
    }
};

/* 点p在有向直线L的左边 */
bool OnLeft(Line L, Point p) {
    return dcmp(Cross(L.v, p - L.p)) > 0; 
}

/* 二直线交点，假设交点唯一存在。*/
Point GLI(Line a, Line b) {
    Point u = a.p - b.p;
    double t = Cross(b.v, u) / Cross(a.v, b.v);
    return a.p + a.v * t;
}

int HPI(Line* L, int n, Point* Q) {
    sort(L, L + n); /// 极角排序

    int st, ed; /// 双端队列的第一个元素和最后一个元素的下标

    Point *p = new Point[n]; /// p[i]为q[i]和q[i+1]的交点
    Line *q = new Line[n]; /// 双端队列
    q[st = ed = 0] = L[0];

    rep(i, 1, n - 1) {
        while(st < ed && !OnLeft(L[i], p[ed - 1])) ed--;
        while(st < ed && !OnLeft(L[i], p[st])) st++;

        q[++ed] = L[i];

        /// 平行取内测那条
        if(fabs(Cross(q[ed].v, q[ed - 1].v)) < eps) {
            ed--;
            if(OnLeft(q[ed], L[i].p)) q[ed] = L[i];
        }

        if(st < ed) p[ed - 1] = GLI(q[ed - 1], q[ed]);

    }

    while(st < ed && !OnLeft(q[st], p[ed - 1])) ed--;

    if(ed - st <= 1) return 0;

    p[ed] = GLI(q[ed], q[st]);

    int m = 0;
    rep(i, st, ed) Q[m++] = p[i];
    return m;
}

Point P[N], Q[N],tmpa, tmpb;

Line L[N];

int n;
double a[N], b[N], c[N], A, B, C;

bool judge(int x) {

    int cnt = 4;
    
     ///给半平面加一个框，这样可以使解x，y都大于0，也可以避免所有半平面交起来后为不为凸多边形，而是一个敞开的区域
    ///如果题目输入的不是一个多边形，而是本题这种输入若干不等式组的情况，这样的限定就是必须的，不然有bug，例如，两条线是平行的(但是极角不同)，
    ///极角排序后又挨在一起, 那么就可能求它们的交点，就容易出错
    tmpa.x = 0; tmpa.y = 0; tmpb.x = maxL; tmpb.y = 0;
    L[0] = Line(tmpa, tmpb - tmpa);
    tmpa = tmpb; tmpb.x = maxL; tmpb.y = maxL;
    L[1] = Line(tmpa, tmpb - tmpa);
    tmpa = tmpb; tmpb.x = 0;
    L[2] = Line(tmpa, tmpb - tmpa);
    tmpa = tmpb; tmpb.y = 0;
    L[3] = Line(tmpa, tmpb - tmpa);
    
    rep(i, 1, n) {
        if(i == x) continue;
        A = 1.0 / a[i] - 1.0 / a[x];
        B = 1.0 / b[i] - 1.0 / b[x];
        C = 1.0 / c[i] - 1.0 / c[x];

        int d1 = dcmp(A), d2 = dcmp(B), d3 = dcmp(C);
        
        /* 下面是根据a*x+b*y+c>0取向量p1p2，
           其中p1（x1，y1），p2（x2，y2）
           就是将直线转化为以两点的表示，取向量p1p2左半为半平面
        */
        if(!d1) {
            if(!d2) {
                if(d3 <= 0) return false;
                continue;
            }
            tmpa.x = 0;
            tmpb.x = d2;
            tmpa.y = tmpb.y = -C / B;
        }
        else {
            if(!d2) {
                tmpa.x = tmpb.x = -C / A;
                tmpa.y = 0; tmpb.y = -d1;
            }
            else {
                tmpa.x = 0; tmpa.y = -C / B;
                tmpb.x = d2;
                tmpb.y = -(C + A * tmpb.x) / B;
            }
        }

        L[cnt++] = Line(tmpa, tmpb - tmpa);

    }

    if(HPI(L, cnt, Q) == 0) return false;
    else return true;

}

void solve() {
    
    scanf("%d", &n);
    rep(i, 1, n) scanf("%lf %lf %lf", &a[i], &b[i], &c[i]);

    rep(i, 1, n) {
        if(judge(i)) puts("Yes");
        else puts("No");
    }
}

int main() {

//    int _; scanf("%d", &_);
//    while(_--) solve();

    solve();

    return 0;
}