hdoj4717The Moving Points【三分】

B - The Moving Points

Time Limit:3000MS     Memory Limit:32768KB     64bit IO Format:%I64d & %I64u

Submit Status Practice HDU 4717

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

220 0 1 02 0 -1 020 0 1 02 1 -1 0 

 

Sample Output

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

提议给出一些点和他们的速度求使他们任意两点间的距离的最大值最小

#include<cstdio>#include<cstdlib>#include<cstring>#include<cmath>#include<algorithm>#define eps 1e-8using namespace std;struct point{double x,y,vx,vy;}A[310];int n;double MAX(double a,double b){return a>b?a:b;}double MIN(double a,double b){return a<b?a:b;}double dist(point a,point b){return sqrt((a.x-b.x)*(a.x-b.x)+(a.y-b.y)*(a.y-b.y));}double distmin(double t){double max=0;for(int i=0;i<n;++i){for(int j=i+1;j<n;++j){point a,b;a.x=A[i].x+t*A[i].vx;a.y=A[i].y+t*A[i].vy;b.x=A[j].x+t*A[j].vx;b.y=A[j].y+t*A[j].vy;max=MAX(max,dist(a,b));}}return max;}int main(){int t,i,j,k=1;scanf("%d",&t);while(t--){scanf("%d",&n);for(i=0;i<n;++i){scanf("%lf%lf%lf%lf",&A[i].x,&A[i].y,&A[i].vx,&A[i].vy);}int size=100;double left=0,right=10000;while(size--){double mid=(left+right)/2;double mmid=(mid+right)/2;if(distmin(mid)<distmin(mmid)){right=mmid;}else {left=mid;}}printf("Case #%d: %.2lf %.2lf\n",k++,left,distmin(left));}return 0;}