POJ 2074 Line of Sight
2013-08-19 20:18
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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。
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_JUDGE freopen("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; }
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