HDU 5027 Help!

三分趣题, 分类讨论.

之前C++读入被各种卡时间= = 真是坑, G++就过了

计算几何什么的就别卡时间啦...

P.S. 为何网上有人用圆与线段交(疑惑)?

附上奇怪的代码

#include<iostream>#include<cstdio>#include<complex>#include<vector>#include<cmath>#include<cstdlib>#include<algorithm>using namespace std;const double eps=1.0e-8;typedef complex<double> point;typedef vector<point> polygon;struct line : public vector<point> {    line() {}    line(const point& a, const point& b) { push_back(a); push_back(b); }};inline point vec(const line& l) { return l[1]-l[0];}inline double dot  (const point& a, const point& b) { return (a*conj(b)).real();}inline double cross(const point& a, const point& b) { return (conj(a)*b).imag();}inline int dlcmp(const double& x){    return x<-eps?-1:x>eps;}bool intersectSP(const line& s, const point& p) {    return abs(s[0]-p)+abs(s[1]-p)<abs(s[1]-s[0])+eps;}point projection(const line &l, const point &p) {    double t = dot(p-l[0], l[0]-l[1]) / norm(l[0]-l[1]);    return l[0] + t*(l[0]-l[1]);}double distanceSP(const line &s, const point &p) {    const point& r = projection(s, p);    if (intersectSP(s, r)) return abs(r - p);    return min(abs(s[0]-p), abs(s[1]-p));}polygon p;const int MAXN=50005;bool direction[MAXN];inline double calc(const point& start, const point& m, const point& target, const double& Vr, const double& Vs){    return abs(start-m)/Vr+abs(m-target)/Vs;}point m1, m2, v;double mid, midd;double solve(const line& S, const point& start, const point& target, const double& Vr, const double& Vs, double& ts){    double l=0.0, r=1.0;    v=vec(S);    while(l+eps<r){        mid=(l+r)/2.0;        midd=(mid+r)/2.0;        m1=S[0]+mid*v;        m2=S[0]+midd*v;        if(calc(start,m1,target,Vr,Vs)<calc(start,m2,target,Vr,Vs)){            r=midd;        }        else{            l=mid;        }    }    ts=abs(S[0]+l*v-target)/Vs;    return calc(start,S[0]+l*v,target,Vr,Vs);}point cut[2];int R[2];point p1, p2;int main(){    int T, n; cin>>T;    while(T--){        double Ts, Vr, Vs; cin>>Ts>>Vr>>Vs;        double Xo, Yo, Xp, Yp; cin>>Xo>>Yo>>Xp>>Yp;        point O(Xo,Yo), P(Xp,Yp);        cin>>n;        p.clear();        double x, y;        for(int i=0; i<n; i++){            scanf("%lf%lf",&x,&y);            p.push_back(point(x,y));        }        if(dlcmp(cross(p[1]-p[0],p[2]-p[1]))<0){            reverse(p.begin(),p.end());        }        double carry_and_swim=1.0e20;        for(int i=0; i<n; i++){            carry_and_swim=min(carry_and_swim,distanceSP(line(p[i],p[(i+1)%n]),P));        }        carry_and_swim=2.0*carry_and_swim/Vs;        double sum=0.0;        for(int i=0; i<n; i++){            direction[i]=dlcmp(cross(p[i]-O,p[(i+1)%n]-p[i]))>=0;            sum+=abs(p[i]-p[(i+1)%n]);        }        int cnt=0;        for(int i=0; i<n; i++){            if(direction[i]!=direction[(i+1)%n]){                R[cnt++]=(i+1)%n;            }            if(intersectSP(line(p[i],p[(i+1)%n]),O)){                cut[0]=p[i];                cut[1]=p[(i+1)%n];                R[0]=i; R[1]=(i+1)%n;            }        }        if(cnt!=0){            if(direction[R[0]]!=0) swap(R[0],R[1]);            cut[0]=p[R[0]]; cut[1]=p[R[1]];        }        double ans=1.0e20;        int st=R[0], ed=R[1];        if(ed<st) ed+=n;        double ft=0.0;        double ts, ts1, ts2, ret;        for(int i=st; i<ed; i++){            p1=p[i%n], p2=p[(i+1)%n];            ret=solve(line(p1,p2),O,P,Vr,Vs,ts);            if(ts+carry_and_swim-eps<=Ts){                ans=min(ans,ret);            }            ft+=abs(p1-p2);        }        swap(st,ed);        if(ed<st) ed+=n;        double bk=0.0, al=(sum-ft), d1, d2;        for(int i=st; i<ed; i++){            p1=p[i%n], p2=p[(i+1)%n];            d1=(abs(O-cut[1])+bk)/Vr+solve(line(p1,p2),p1,P,Vr,Vs,ts1);            d2=(abs(O-cut[0])+al-bk-abs(p1-p2))/Vr+solve(line(p2,p1),p2,P,Vr,Vs,ts2);            if(ts1+carry_and_swim-eps<=Ts){                ans=min(ans,d1);            }            if(ts2+carry_and_swim-eps<=Ts){                ans=min(ans,d2);            }            bk+=abs(p1-p2);        }        ans+=carry_and_swim;        if(ans<=1.0e19) printf("%.2f\n",ans);        else printf("-1\n");    }    return 0;}