hdu 3400 Line belt

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先求一个点到一条线段的最小值。

从P 到B .

从 c 到 A 肯定是增加的。

设 cd 为x ，cb 是 L2 pc 是L1

线段上速率是 k2 其余为 k1

t = sqrt(L1*L1 + x*x )/k1 + (L1 -x);

求导。

f = 2/(k1 * sqrt(L1 *L1 + x*x)) *x - 1/k2

随着x 的增大 f 逐渐变大。但是由于 k1 k2 不确定所以 f 可能由负变成正或者只有正。

加上AC 同样满足三分性质。

x = L1 / sqrt( (4*K2*k2 - K1*K1)/ (K1*K1))

所以 x 跟 L1 成正比（重要）。

在根据成正比这个性质很容易分析出两条线的时候也是满足三分性质的（同样跟垂线有关，以此为分界线。趋近B端的变化成以一定的比率上升或下降，在加上A到垂足部分就是可能出现下凸。）

#include <vector>#include <list>#include <map>#include <set>#include <deque>#include <stack>#include <cstring>#include <bitset>#include <algorithm>#include <functional>#include <numeric>#include <utility>#include <sstream>#include <iostream>#include <iomanip>#include <cstdio>#include <cmath>#include <cstdlib>#include <ctime>#include <assert.h>#include <queue>#define REP(i,n) for(int i=0;i<n;i++)#define TR(i,x) for(typeof(x.begin()) i=x.begin();i!=x.end();i++)#define ALLL(x) x.begin(),x.end()#define SORT(x) sort(ALLL(x))#define CLEAR(x) memset(x,0,sizeof(x))#define FILLL(x,c) memset(x,c,sizeof(x))using namespace std;const double eps = 1e-9;#define LL long long #define pb push_backconst int maxn  = 11000;double ax , ay, bx,by,cx,cy,dx,dy;double p ,q ,r;double s(double x1 ,double y1 ,double x2,double y2){     return sqrt((x2- x1)*(x2 -x1) + (y2 - y1)*(y2 -y1));}double f3(double a,double x,double y){ double x2  = cx + (dx- cx)*a;     double y2  = cy + (dy-cy)*a;     return s(x,y,x2,y2)/r +  s(x2,y2,dx,dy)/q;}double f2(double x ,double y){  double left = 0;  double right =1;  while(right - left>1e-6){   double  lmid = (left +right )/2;   double  rmid = (lmid + right)/2;    if(f3(lmid,x,y)>f3(rmid,x,y)){      left = lmid;    }else{     right = rmid;    }  //  cout << lmid << "  "<<rmid <<endl;  }     return f3(left,x,y);}double f(double a){ double x = ax + (bx-ax)*a; double y = ay + (by-ay)*a; double ans = s(ax,ay,x,y)/p; ans += f2(x,y); return ans;}void solve(){    double left = 0;    double right =1;    while(right - left > 1e-6){    double lmid = (left +right)/2;    double rmid = (lmid +right)/2;    if(f(lmid)>f(rmid)){            left = lmid;    }else{        right = rmid;    }        }    double ans = f(left);    printf("%.2f\n",ans);}int main(){    int  t;    cin >>t ;    while(~scanf("%lf%lf%lf%lf",&ax,&ay,&bx,&by),t--){      scanf("%lf%lf%lf%lf",&cx,&cy,&dx,&dy);      scanf("%lf%lf%lf",&p,&q,&r);      solve();    //  cout << f3(1,0,0)<<endl;    }    return 0;}