hdu4717(三分法)

The Moving Points

Time Limit: 6000/3000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others)
Total Submission(s): 937    Accepted Submission(s): 380

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

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

 

Source

2013 ACM/ICPC Asia Regional Online —— Warmup2

 

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           本题给定一些点，然后个顶点的速度，求所有点对的最大距离最小的时刻和这个最小距离。

 

           题目暗示这样的时刻唯一，所以可以这只三分求极值。时间复杂度O(N^2*log3(N))，可以接受。

#include<iostream>#include<cmath>using namespace std;const int MAXN=300+100;const double eps=1e-8;int n;double Time;struct point{double x,y;double s_x,s_y;}MyPoint[MAXN];double GetMinDistence(double tim){double tMinD=0;for(int i=0;i<n;i++){for(int j=i+1;j<n;j++){double tix=MyPoint[i].x+MyPoint[i].s_x*tim;double tiy=MyPoint[i].y+MyPoint[i].s_y*tim;double tjx=MyPoint[j].x+MyPoint[j].s_x*tim;double tjy=MyPoint[j].y+MyPoint[j].s_y*tim;double tmp=sqrt((tix-tjx)*(tix-tjx)+(tiy-tjy)*(tiy-tjy));if(tmp>tMinD)tMinD=tmp;}}return tMinD;}void Solve(){double low,high,mid,midmid;low=0.0,high=1e10;while(high-low>=eps){mid=(low+high)/2.0;midmid=(mid+high)/2.0;if(GetMinDistence(midmid)<GetMinDistence(mid))low=mid;else high=midmid;}Time=(high+low)/2;}int main(){int cas,tag=0;cin>>cas;while(cas--){scanf("%d",&n);for(int i=0;i<n;i++)scanf("%lf%lf%lf%lf",&MyPoint[i].x,&MyPoint[i].y,&MyPoint[i].s_x,&MyPoint[i].s_y);Solve();printf("Case #%d: %0.2lf %0.2lf\n",++tag,Time,GetMinDistence(Time));}system("pause");return 0;}