hdu 4717 n个点距离

传送门

The Moving Points

Time Limit: 6000/3000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others)
Total Submission(s): 2024    Accepted Submission(s): 836

Problem Description

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 ^th point, 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
  

 

Source

2013 ACM/ICPC Asia Regional Online —— Warmup2

 

Recommend

zhuyuanchen520

题意：给N个点，给出N个点的方向和移动速度，求每个时刻N个点中任意两点的最大值中的最小值，以及取最小值的时刻

每组点和速度，根据变量t，横纵坐标就是关于t的以为函数。   求两点距离的平方就是下凸函数。fi(x)=ai*t*t+bi*t+ci; 而求n*(n-1)/2个距离的最大值，s(x)=max{fi(x)},也是下凸函数。

//china no.1
#include <vector>
#include <iostream>
#include <string>
#include <map>
#include <stack>
#include <cstring>
#include <queue>
#include <list>
#include <stdio.h>
#include <set>
#include <algorithm>
#include <cstdlib>
#include <cmath>
#include <iomanip>
#include <cctype>
#include <sstream>
#include <functional>
using namespace std;

#define pi acos(-1)
#define endl '\n'
#define srand() srand(time(0));
#define me(x) memset(x,0,sizeof(x));
#define foreach(it,a) for(__typeof((a).begin()) it=(a).begin();it!=(a).end();it++)
#define close() ios::sync_with_stdio(0); cin.tie(0);
typedef long long LL;
const int INF=0x3f3f3f3f;
const LL LINF=0x3f3f3f3f3f3f3f3fLL;
//const int dx[]={-1,0,1,0,-1,-1,1,1};
//const int dy[]={0,1,0,-1,1,-1,1,-1};
const int maxn=1e3+5;
const int maxx=1e5+100;
const double EPS=1e-7;
const int MOD=10000007;
#define mod(x) ((x)%MOD);
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));}

#define FOR(x,n,i) for(int i=x;i<=n;i++)
#define FOr(x,n,i) for(int i=x;i<n;i++)
#define W while
#define sgn(x) ((x) < 0 ? -1 : (x) > 0)

int n;
struct node
{
    double x,y,vx,vy;
}Q[maxx];
double cross(node a,node b,double t)
{
    return (a.x+a.vx*t-b.x-b.vx*t)*(a.x+a.vx*t-b.x-b.vx*t)+(a.y+a.vy*t-b.y-b.vy*t)*(a.y+a.vy*t-b.y-b.vy*t);
}
double calc(double x)
{
    double ans=0;
    FOR(1,n,i)
        FOR(i+1,n,j)
            ans=max(ans,cross(Q[i],Q[j],x));
    //cout<<ans<<endl;
    return ans;
}
int main()
{
    int t;
    cin>>t;
    FOR(1,t,cas)
    {
        cin>>n;
        FOR(1,n,i)
            cin>>Q[i].x>>Q[i].y>>Q[i].vx>>Q[i].vy;
        double l=0,r=1e8;
        int len=300;
        W(len--)
        {
            double mid=l+(r-l)/3;
            double midmid=r-(r-l)/3;
            if(calc(mid)<calc(midmid)) r=midmid;
            else l=mid;
        }
       // printf("%.2lf\n",sqrt(calc(l)));
        printf("Case #%d: %.2lf %.2lf\n",cas,l,sqrt(calc(l)));
    }
}