【三分入门】HDU - 4717 Q - The Moving Points

Q - The Moving Points  HDU - 4717

There are N points in total. Every point moves in certain direction and certain speed. We want to know at what time that the largest distance between any two points would be minimum. And also, we require you to calculate that minimum distance. We guarantee that no two points will move in exactly same speed and direction.

Input

The rst line has a number T (T <= 10) , indicating the number of test cases. 
For each test case, first line has a single number N (N <= 300), which is the number of points. 
For next N lines, each come with four integers X i, Y i, VX i and VY i (-10 6 <= X i, Y i <= 10 6, -10 2 <= VX i , VY i <= 10 2), (X i, Y i) is the position of the i thpoint, and (VX i , VY i) is its speed with direction. That is to say, after 1 second, this point will move to (X i + VX i , Y i + VY i).

Output

For test case X, output "Case #X: " first, then output two numbers, rounded to 0.01, as the answer of time and distance.

Sample Input

2
2
0 0 1 0
2 0 -1 0
2
0 0 1 0
2 1 -1 0

Sample Output

Case #1: 1.00 0.00
Case #2: 1.00 1.00

给你n个点的坐标和横纵坐标每秒变换的值，问你哪个时间这n个点中的最大距离最小值

因为只有300个点，又知道每两个点的距离肯定是成先减后增的抛物线，所以我们三分顶点

每次都n^2找最大值，如果比原来求得的更小就更新时间和答案

先减后增的抛物线
int SanFen(int l,int r) //找凸点  
{  
    while(l < r-1)  
    {  
        int mid  = (l+r)/2;  
        int mmid = (mid+r)/2;  
        if( f(mid) > f(mmid) )  
            l = mid;  
        else  
            r = mmid;  
    }  
    return f(l) > f(r) ? l : r;  
}  

先增后减的抛物线
int SanFen(int l,int r) //找凸点  
{  
    while(l < r-1)  
    {  
        int mid  = (l+r)/2;  
        int mmid = (mid+r)/2;  
        if( f(mid) > f(mmid) )  
            r = mmid;  
        else  
            l = mid;  
    }  
    return f(l) > f(r) ? l : r;  
}  

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#include <bits/stdc++.h>
using namespace std;
#define eps 1e-4
int n;
struct node
{
    double x,y,vx,vy;
}p[350];

double cal(int a,int b,double t)
{
    double x1=p[a].x+t*p[a].vx,x2=p[b].x+t*p[b].vx;
    double y1=p[a].y+t*p[a].vy,y2=p[b].y+t*p[b].vy;
    return sqrt((x1-x2)*(x1-x2)+(y1-y2)*(y1-y2));
}

double check(double t)
{
    double ans=0.0;
    for(int i=1;i<=n;i++)
    {
        for(int j=i+1;j<=n;j++)
        {
            ans=max(ans,cal(i,j,t));
        }
    }
    return ans;
}

int main()
{
    int T;
    scanf("%d",&T);
    int cas=0;
    while(T--)
    {
        scanf("%d",&n);
        for(int i=1;i<=n;i++)
        {
            scanf("%lf%lf%lf%lf",&p[i].x,&p[i].y,&p[i].vx,&p[i].vy);
        }
        double l=0.0,r=100000000000.0,ans_time=0.0,ans_path=10000000000000;
        while(l+eps<=r)
        {
            double mid=(l+r)/2;
            double midd=(mid+r)/2;
            if(check(mid)<=check(midd)) r=midd;
            else l=mid;
            if(ans_path>check(l))
            {
                ans_time=l;
                ans_path=check(l);
            }
        }
        printf("Case #%d: %.2f %.2f\n",++cas,ans_time,ans_path);
    }
    return 0;
}