The Moving Points -HDU 4717 三分

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提比-我有特殊的AC技巧

最新推荐文章于 2024-02-22 21:41:05 发布

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本文介绍了一道算法题目，要求通过三分法找到使多个移动点间的最大距离达到最小值的时间点及其具体距离。该问题可通过计算各点移动速度及位置来解决。

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The Moving Points

Time Limit: 6000/3000 MS (Java/Others) Memory Limit: 32768/32768 K (Java/Others)
Total Submission(s): 1192 Accepted Submission(s): 483

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 (-10⁶ <= X_i, Y_i <= 10⁶, -10² <= VX_i , VY_i <= 10²), (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

Sample Output

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

题意：找到一个时刻使每两个点的距离的最大值最小。

思路：三分这个时间，然后找到最小的最大值，因为这个值是一个二次函数，所以要用到三分。

AC代码如下：

#include<cstdio>
#include<cstring>
#include<algorithm>
#include<cmath>
using namespace std;
struct node
{ double x,y,vx,vy;
}point[310];
int n;
double mi,mi2,ti,eps=1e-8;

double dis(int i,int j,double ti)
{ return sqrt( (point[i].x+point[i].vx*ti-point[j].x-point[j].vx*ti)*(point[i].x+point[i].vx*ti-point[j].x-point[j].vx*ti)
              +(point[i].y+point[i].vy*ti-point[j].y-point[j].vy*ti)*(point[i].y+point[i].vy*ti-point[j].y-point[j].vy*ti) );
}
double solve(double ti)
{ double ans=0;
  int i,j;
  for(i=1;i<n;i++)
   for(j=i+1;j<=n;j++)
    ans=max(ans,dis(i,j,ti));
  return ans;
}
int main()
{ int T,t,i,j;
  double l=0,r=1000000000,k;
  scanf("%d",&T);
  for(t=1;t<=T;t++)
  { scanf("%d",&n);
    for(i=1;i<=n;i++)
     scanf("%lf%lf%lf%lf",&point[i].x,&point[i].y,&point[i].vx,&point[i].vy);
    l=0,r=1000000000;
    while(r-l>eps)
    { k=(r-l)/3;
      mi=l+k;mi2=l+k*2;
      if(solve(mi)<solve(mi2))
       r=mi2;
      else
       l=mi;
    }
    printf("Case #%d: %.2f %.2f\n",t,l,solve(l));
  }
}