• HDU 3400 Line belt (三分法)


    Line belt

    Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others)
    Total Submission(s): 1820    Accepted Submission(s): 683


    Problem Description
    In a two-dimensional plane there are two line belts, there are two segments AB and CD, lxhgww's speed on AB is P and on CD is Q, he can move with the speed R on other area on the plane.
    How long must he take to travel from A to D?
     
    Input
    The first line is the case number T.
    For each case, there are three lines.
    The first line, four integers, the coordinates of A and B: Ax Ay Bx By.
    The second line , four integers, the coordinates of C and D:Cx Cy Dx Dy.
    The third line, three integers, P Q R.
    0<= Ax,Ay,Bx,By,Cx,Cy,Dx,Dy<=1000
    1<=P,Q,R<=10
     
    Output
    The minimum time to travel from A to D, round to two decimals.
     
    Sample Input
    1 0 0 0 100 100 0 100 100 2 2 1
     
    Sample Output
    136.60
     
    Author
    lxhgww&&momodi
     
    Source
     
    Recommend
    lcy
     
     
     
    神奇的三分法。
    就是利用凸性。
    #include<stdio.h>
    #include<iostream>
    #include<math.h>
    #include<iostream>
    #include<algorithm>
    #include<string.h>
    using namespace std;
    const double eps=1e-5;
    struct point
    {
        double x,y;
    };
    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 p,q,r;
    
    double find2(point a,point c,point d)
    {
        point left,right;
        point mid,midmid;
        double t1,t2;
        left=c;right=d;
        do
        {
            mid.x=(left.x+right.x)/2;
            mid.y=(left.y+right.y)/2;
            midmid.x=(mid.x+right.x)/2;
            midmid.y=(mid.y+right.y)/2;
            t1=dis(a,mid)/r+dis(mid,d)/q;
            t2=dis(a,midmid)/r+dis(midmid,d)/q;
            if(t1>t2)left=mid;
            else right=midmid;
        }
        while(dis(left,right)>=eps);
        return t1;
    }
    double find(point a,point b,point c,point d)
    {
        point left,right;
        point mid,midmid;
        double t1,t2;
        left=a;
        right=b;
        do
        {
            mid.x=(left.x+right.x)/2;
            mid.y=(left.y+right.y)/2;
            midmid.x=(mid.x+right.x)/2;
            midmid.y=(mid.y+right.y)/2;
            t1=dis(a,mid)/p+find2(mid,c,d);
            t2=dis(a,midmid)/p+find2(midmid,c,d);
            if(t1>t2)left=mid;
            else right=midmid;
        }while(dis(right,left)>=eps);
        return t1;
    }
    int main()
    {
        int T;
        point a,b,c,d;
        scanf("%d",&T);
        while(T--)
        {
            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);
            printf("%.2lf\n",find(a,b,c,d));
        }
        return 0;
    }
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  • 原文地址:https://www.cnblogs.com/kuangbin/p/2667319.html
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