hdu 4717(三分) The Moving Points

链接:http://acm.hdu.edu.cn/showproblem.php?pid=4717

n个点，给出初始坐标和移动的速度和移动的方向，求在哪一时刻任意两点之间的距离的最大值的最小。

对于最大值的最小问题首先就会想到二分，然而由于两点之间的距离是呈抛物线状，所以三分的方法又浮上脑海，当然二分也可以做，不过思维上更复杂一些

 #include<cmath>
 #include<cstdio>
 #include<cstring>
 #include<iostream>
 #include<algorithm>
 using namespace std;

 const double eps = 0.000001;
 const int M = ;
 int n;

 struct point{
     double x,y,vx,vy;
     void read(){scanf("%lf%lf%lf%lf",&x,&y,&vx,&vy);}
 }po[M];

 double max(double x,double y){return x>y?x:y;}

 double dis(double x1,double y1,double x2,double y2)
 {
     return sqrt((x1-x2)*(x1-x2)+(y1-y2)*(y1-y2));
 }

 double cal(double t)
 {
     double ans=;
     for (int i= ; i<=n ; i++){
         for (int j=i+ ; j<=n ; j++){
             double num=dis(po[i].x+t*po[i].vx,po[i].y+t*po[i].vy,po[j].x+t*po[j].vx,po[j].y+t*po[j].vy);
             ans=max(num,ans);
         }
     }
     return ans;
 }

 double solve(double l,double r)
 {
     while (r-l>=eps)
     {
         double mid1=(l*+r)/;
         double mid2=(l+r*)/;
         if (cal(mid2)-cal(mid1)>eps)
             r=mid2;
         else
             l=mid1;
     }
     return l;
 }

 int main()
 {
     int t,cas=;
     scanf("%d",&t);
     while (t--)
     {
         scanf("%d",&n);
         for (int i= ; i<=n ; i++)
             po[i].read();
         double ans=solve(,1e8);
         printf("Case #%d: ", ++cas);
         if (n==) printf("0.00 0.00\n");
         else printf("%.2lf %.2lf\n",ans,cal(ans));
     }
     return ;
 }