【三分】hdu 3400

題意：給出兩條平行線AB，CD，AB速度為p，CD速度為q，線段以外為r，問從A到D最短時間？

方法：三分。簡單分析後發現這個函數是單峰函數，先減後增，三分適用；

實現：先對AB三分，裡面嵌套著對CD三分；

#include <list>#include <map>#include <set>#include <queue>#include <string>#include <deque>#include <stack>#include <algorithm>#include <iostream>#include <iomanip>#include <cstdio>#include <cmath>#include <cstdlib>#include <limits.h>#include <time.h>#include <string.h>using namespace std;#define LL long long#define PI acos(-1.0)#define MAX INT_MAX#define MIN INT_MIN#define eps 1e-10#define FRE freopen("a.txt","r",stdin)#define N 1000struct point{    double x,y;};double p,q,r;double dis(point a,point b){    return sqrt( (a.x-b.x) * (a.x-b.x) + (a.y-b.y) * (a.y-b.y) );}double cal_cd(point c,point d,point mm){    int i,j;    point ll,rr,mid,midmid;    ll=c;    rr=d;    double t1,t2;    do{        mid.x=(ll.x+rr.x)/2.0;        mid.y=(ll.y+rr.y)/2.0;        midmid.x=(mid.x+rr.x)/2.0;        midmid.y=(mid.y+rr.y)/2.0;        t1=dis(d,mid)/q+dis(mid,mm)/r;        t2=dis(d,midmid)/q+dis(midmid,mm)/r;        if(t1<t2)        rr=midmid;        else        ll=mid;    }while(fabs(t1-t2)>=eps);    return t1;}double cal_ab(point a,point b,point c,point d){    int i,j;    point ll,rr,mid,midmid;    ll=a;    rr=b;    double t1,t2;    do{        mid.x=(ll.x+rr.x)/2.0;        mid.y=(ll.y+rr.y)/2.0;        midmid.x=(mid.x+rr.x)/2.0;        midmid.y=(mid.y+rr.y)/2.0;        double tmp1=cal_cd(c,d,mid);        double tmp2=cal_cd(c,d,midmid);        t1=dis(a,mid)/p+tmp1;        t2=dis(a,midmid)/p+tmp2;        if(t1<t2)           rr=midmid;        else        ll=mid;    }while(fabs(t1-t2)>=eps);    return t1;}int main(){    int t;    scanf("%d",&t);    while(t--){        point a,b,c,d;        scanf("%lf %lf %lf %lf",&a.x,&a.y,&b.x,&b.y);        scanf("%lf %lf %lf %lf",&c.x,&c.y,&d.x,&d.y);        scanf("%lf %lf %lf",&p,&q,&r);        double ans=cal_ab(a,b,c,d);        printf("%.2f\n",ans);    }    return 0;}