poj2074 Line of Sight

计算几何题目，特别要注意障碍物位于House Line上方或者Property Line下方时要首先排除。

http://poj.org/problem?id=2074

#include <cstdio>
#include <cstring>
#include <algorithm>
#include <queue>

using namespace std;
const int maxn = 1e6;

double Hx1, Hx2, Hy, Px1, Px2, Py;
struct Point{
    double x1, x2;
    Point(double x1 = 0, double x2 = 0) : x1(x1), x2(x2) {}
    bool operator < (const Point& rhs) const{
        return x1 < rhs.x1 || (x1 == rhs.x1 && x2 < rhs.x2);
    }
    Point operator - (const Point& rhs) const{
        return Point(x1 - rhs.x1, x2 - rhs.x2);
    }
};
Point s[maxn];
int n, k;
double x1, x2, y;

double Cross(Point a, Point b)    { return a.x1 * b.x2 - b.x1 * a.x2; }

double getPosR(double x, double y){
    Point vect1 = Point(Hx1, Hy) - Point(x, y);
    Point vect2 = Point(Px2, Py) - Point(Hx1, Hy);
    if(Cross(vect1, vect2) >= 0) return Px2;
    Point vect3 = Point(Px1, Py) - Point(Hx1, Hy);
    if(Cross(vect1, vect3) <= 0) return Px1;
    return (x - Hx1) * (Py - Hy) / (y - Hy) + Hx1;
}

double getPosL(double x, double y){
    Point vect1 = Point(Hx2, Hy) - Point(x, y);
    Point vect2 = Point(Px2, Py) - Point(Hx2, Hy);
    if(Cross(vect1, vect2) >= 0) return Px2;
    Point vect3 = Point(Px1, Py) - Point(Hx2, Hy);
    if(Cross(vect1, vect3) <= 0) return Px1;
    return (x - Hx2) * (Py - Hy) / (y - Hy) + Hx2;
}

int main(){
    freopen("in.txt", "r", stdin);
    while(~scanf("%lf%lf%lf", &Hx1, &Hx2, &Hy) && (Hx1 + Hx2 + Hy)){
        scanf("%lf%lf%lf", &Px1, &Px2, &Py);
        scanf("%d", &n);
        k = 0;
        for(int i = 0; i < n; i++){
            scanf("%lf%lf%lf", &x1, &x2, &y);
            if(y >= Hy || y <= Py) continue;
            double tx1 = getPosL(x1, y);
            double tx2 = getPosR(x2, y);
            if(tx1 != tx2) s[k++] = Point(tx1, tx2);
        }
        if(!k){
            printf("%.2f
", Px2 - Px1);
            continue;
        }
        sort(s, s + k);
        double ans = s[0].x1 - Px1;
        for(int i = 0; i < k; i++){
            double rear = s[i].x2;
            int j = i;
            while(i + 1 < k && s[i + 1].x2 <= rear) ++i;
            if(i == k - 1){
                ans = max(ans, Px2 - rear);
                break;
            }
            double front = s[i + 1].x1;
            if(front > rear) ans = max(ans, front - rear);
        }
        if(!ans) puts("No View");
        else printf("%.2f
", ans);
    }
    return 0;
}