HDU3400（三分）

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思路：设a为AB上一点，d为CD上一点，分别对a到A的距离和d到D的距离三分（对应的时间的变化均为凹函数），每次先枚举a的距离，再对d到D 的距离三分出此时对应的最小时间，直到a到A的距离枚举完，得到最小时间。

#include<cstdio>#include<iostream>#include<cstring>#include<algorithm>#include<cmath>#define eps 1e-8using namespace std;struct node{    double x,y;};node A,B,C,D;double P,Q,R;double ab,cd;double dist(node a, node b){    return sqrt(eps+((a.x-b.x)*(a.x-b.x)+(a.y-b.y)*(a.y-b.y)));}double ansCD(double tmp){    node a,d;    a.x=A.x+(B.x-A.x)/ab*tmp,a.y=A.y+(B.y-A.y)/ab*tmp;    double mid,midmid;    double low=0, high=dist(C,D);    while(low+eps<high)    {         mid=(low+high)/2;         d.x=D.x+(C.x-D.x)/cd*mid,d.y=D.y+(C.y-D.y)/cd*mid;         double cmid=dist(a,d)/R+mid/Q;         midmid=(mid+high)/2;         d.x=D.x+(C.x-D.x)/cd*midmid,d.y=D.y+(C.y-D.y)/cd*midmid;         double cmidmid=dist(a,d)/R+midmid/Q;         if(cmid>cmidmid)            low=mid;         else            high=midmid;    }    return dist(a,d)/R+midmid/Q;}double ansAB(){    double mid,midmid;    double low=0,high=dist(A,B);    while(low+eps<high)    {        mid=(low+high)/2;        midmid=(mid+high)/2;        double cmid=mid/P+ansCD(mid);        double cmidmid=midmid/P+ansCD(midmid);        if(cmid>cmidmid)            low=mid;        else if(cmid<cmidmid)            high=midmid;            else if(cmid==cmidmid)            break;    }    return mid/P+ansCD(mid);}int main(){    int T;    scanf("%d",&T);    while(T--)    {        node tmp;        tmp.x=0,tmp.y=0;        scanf("%lf%lf%lf%lf%lf%lf%lf%lf",&A.x,&A.y,&B.x,&B.y,&C.x,&C.y,&D.x,&D.y);        scanf("%lf%lf%lf",&P,&Q,&R);        ab=dist(A,B),cd=dist(C,D);        printf("%.2lf\n",ansAB());    }    return  0;}

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