HDU-4717 The Moving Points 三分

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

　　每两个点之间的距离变化是一个单峰函数，因为每次求的值最大值，多个单峰函数的最值函数还是单峰的。。

//STATUS:C++_AC_515MS_256KB
#include <functional>
#include <algorithm>
#include <iostream>
//#include <ext/rope>
#include <fstream>
#include <sstream>
#include <iomanip>
#include <numeric>
#include <cstring>
#include <cassert>
#include <cstdio>
#include <string>
#include <vector>
#include <bitset>
#include <queue>
#include <stack>
#include <cmath>
#include <ctime>
#include <list>
#include <set>
#include <map>
using namespace std;
//#pragma comment(linker,"/STACK:102400000,102400000")
//using namespace __gnu_cxx;
//define
#define pii pair<int,int>
#define mem(a,b) memset(a,b,sizeof(a))
#define lson l,mid,rt<<1
#define rson mid+1,r,rt<<1|1
#define PI acos(-1.0)
//typedef
typedef __int64 LL;
typedef unsigned __int64 ULL;
//const
const int N=310;
const int INF=0x3f3f3f3f;
const int MOD=1000000007,STA=8000010;
const LL LNF=1LL<<60;
const double EPS=1e-5;
const double OO=1e60;
const int dx[4]={-1,0,1,0};
const int dy[4]={0,1,0,-1};
const int day[13]={0,31,28,31,30,31,30,31,31,30,31,30,31};
//Daily Use ...
inline int sign(double x){return (x>EPS)-(x<-EPS);}
template<class T> T gcd(T a,T b){return b?gcd(b,a%b):a;}
template<class T> T lcm(T a,T b){return a/gcd(a,b)*b;}
template<class T> inline T lcm(T a,T b,T d){return a/d*b;}
template<class T> inline T Min(T a,T b){return a<b?a:b;}
template<class T> inline T Max(T a,T b){return a>b?a:b;}
template<class T> inline T Min(T a,T b,T c){return min(min(a, b),c);}
template<class T> inline T Max(T a,T b,T c){return max(max(a, b),c);}
template<class T> inline T Min(T a,T b,T c,T d){return min(min(a, b),min(c,d));}
template<class T> inline T Max(T a,T b,T c,T d){return max(max(a, b),max(c,d));}
//End

struct Node{
    double x,y,dx,dy;
    Node(){}
    Node(int _x,int _y,int _dx,int _dy){x=_x,y=_y,dx=_dx,dy=_dy;}
}nod[N];
int T,n;

double dist(const Node& a,const Node& b,double t)
{
    Node p,q;
    p.x=a.x+a.dx*t,p.y=a.y+a.dy*t;
    q.x=b.x+b.dx*t,q.y=b.y+b.dy*t;
    return (p.x-q.x)*(p.x-q.x)+(p.y-q.y)*(p.y-q.y);
}

double maxdist(double t)
{
    int i,j;
    double hig=0;
    for(i=0;i<n;i++){
        for(j=i+1;j<n;j++){
            hig=Max(hig,dist(nod[i],nod[j],t));
        }
    }
    return hig;
}

double trisection(double l,double r)
{
    double mid,midmid,t1,t2;
    while(r-l>EPS){
        mid=(l+r)/2;
        midmid=(mid+r)/2;
        t1=maxdist(mid);
        t2=maxdist(midmid);
        if(t1<=t2)r=midmid;
        else l=mid;
    }
    return mid;
}

int main(){
 //   freopen("in.txt","r",stdin);
    int ca=1,i,j;
    double x,y,dx,dy,anst;
    scanf("%d",&T);
    while(T--)
    {
        scanf("%d",&n);
        for(i=0;i<n;i++){
            scanf("%lf%lf%lf%lf",&x,&y,&dx,&dy);
            nod[i]=Node(x,y,dx,dy);
        }

        anst=trisection(0,1e6);

        printf("Case #%d: %.2lf %.2lf
",ca++,anst,sqrt(maxdist(anst)));
    }
    return 0;
}