POJ 2074 Line of Sight

POJ 2074 Line of Sight

线段相交

其实就是枚举障碍物的坐端点和房子的右端点   还有就是障碍的右端点和房子的左端点构成的直线和风景的交点对；

然后就是求这中间没有被覆盖的最长的长度。

#include<iostream>#include<cstdio>#include<cstring>#include<algorithm>#include<cmath>#include<vector>using namespace std;#define FOR(i,a,b) for(int (i)=(a);(i)<=(b);(i)++)#define DOR(i,a,b) for(int (i)=(a);(i)>=(b);(i)--)#define oo 1e6#define eps 1e-8#define nMax 100000#define mp make_pair#define pb push_back#define F first#define S second#define bug puts("OOOOh.....");#define zero(x) (((x)>0?(x):-(x))<eps)int dcmp(double x){    if(fabs(x)<eps) return 0;    return x>0?1:-1;}class point {public:    double x,y;    point (double x=0,double y=0):x(x),y(y) {}    void make(double _x,double _y) {x=_x;y=_y;}    void read() { scanf("%lf%lf",&x,&y); }    void out() { printf("%.2lf %.2lf\n",x,y);}    double len() { return sqrt(x*x+y*y); }    point friend operator - (point const& u,point const& v) {        return point(u.x-v.x,u.y-v.y);    }    point friend operator + (point const& u,point const& v) {        return point(u.x+v.x,u.y+v.y);    }    double friend operator * (point const& u,point const& v) {        return u.x*v.y-u.y*v.x;    }    double friend operator ^ (point const& u,point const& v) {        return u.x*v.x+u.y*v.y;    }    point friend operator * (point const& u,double const& k) {        return point(u.x*k,u.y*k);    }};typedef class line{public:    point a,b;    line() {}    line (point a,point b):a(a),b(b){}    void make(point u,point v) {a=u;b=v;}void make(double x1,double x2,double y){make(point(x1,y),point(x2,y));}    void read() { scanf("%lf%lf%lf",&a.x,&b.x,&a.y);b.y=a.y; }friend double intersection(line u,line v);}segment;double intersection(line u,line v){point ret=u.a;double t=(u.a-v.a)*(v.a-v.b)/((u.a-u.b)*(v.a-v.b));ret = ret + (u.b-u.a)*t;return ret.x;}line H,P,s[nMax];int n;vector<pair<double,double> > V;void sovle(){V.clear();for(int i=1;i<=n;i++) {if(dcmp(H.a.y-s[i].a.y)>0 && dcmp(s[i].a.y-P.a.y)>0){V.pb(mp(intersection(line(H.b,s[i].a),P),intersection(line(H.a,s[i].b),P)));}}V.pb(mp(P.b.x,P.b.x));sort(V.begin(),V.end());double ans=0;double nb=P.a.x;for(int i=0;i<V.size();i++) {if(dcmp(V[i].F-nb)>0) {ans = max(ans,V[i].F-nb);}if(dcmp(V[i].S-nb)>0){nb=V[i].S;}}if(dcmp(ans)==0) printf("No View\n");else printf("%.2lf\n",ans + eps);return ;}int main(){#ifndef ONLINE_JUDGEfreopen("input.txt","r",stdin);#endif    double x1,x2,y;while(~scanf("%lf%lf%lf",&x1,&x2,&y),dcmp(x1)||dcmp(x2)||dcmp(y)){H.make(x1,x2,y);P.read();scanf("%d",&n);FOR(i,1,n) s[i].read();sovle();}return 0;}