http://www.pipioj.online/problem.php?id=1025
显然，两个人的距离要么越来越远，要么先越来越近再越来越远，故对时间三分即可求出最短距离。
1 #define bug(x) cout<<#x<<" is "<<x<<endl 2 #define IO std::ios::sync_with_stdio(0) 3 #include <bits/stdc++.h> 4 #define iter ::iterator 5 #define pa pair<int,int> 6 #define pp pair<int,pa> 7 using namespace std; 8 #define ll long long 9 #define mk make_pair 10 #define pb push_back 11 #define se second 12 #define fi first 13 #define ls o<<1 14 #define rs o<<1|1 15 const ll mod=1e9+7; 16 const int N=1e6+10; 17 double eps=1e-10; 18 19 int T; 20 21 double cal(double x1,double y1,double x2,double y2){ 22 return sqrt((x1-x2)*(x1-x2)+(y1-y2)*(y1-y2)); 23 } 24 25 int main(){ 26 scanf("%d",&T); 27 int kase=0; 28 while(T--){ 29 double x1,y1,x2,y2,u1,v1,u2,v2; 30 scanf("%lf%lf%lf%lf",&x1,&y1,&x2,&y2); 31 scanf("%lf%lf%lf%lf",&u1,&v1,&u2,&v2); 32 double ans=1e9; 33 ans=min(ans,cal(x1,y1,x2,y2)); 34 double l=0,r=1e9,nx1,ny1,nx2,ny2; 35 while(r-l>eps){ 36 double d=(r-l)/3; 37 double m1=l+d; 38 double m2=m1+d; 39 40 nx1=x1+m1*u1; 41 ny1=y1+m1*v1; 42 nx2=x2+m1*u2; 43 ny2=y2+m1*v2; 44 double res1=cal(nx1,ny1,nx2,ny2); 45 46 nx1=x1+m2*u1; 47 ny1=y1+m2*v1; 48 nx2=x2+m2*u2; 49 ny2=y2+m2*v2; 50 double res2=cal(nx1,ny1,nx2,ny2); 51 if(res1<res2+eps)r=m2; 52 else l=m1; 53 } 54 nx1=x1+l*u1; 55 ny1=y1+l*v1; 56 nx2=x2+l*u2; 57 ny2=y2+l*v2; 58 ans=min(ans,cal(nx1,ny1,nx2,ny2)); 59 printf("Case %d: %.6lf ",++kase,ans); 60 } 61 }