hdu 4717 The Moving Points（三分）

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http://acm.hdu.edu.cn/showproblem.php?pid=4717

大致题意：给出每个点的坐标以及每个点移动的速度和方向。问在那一时刻点集中最远的距离在所有时刻的最远距离中最小。

比赛时一直以为是计算几何，和线段相交什么的有关。赛后队友说这是道三分，仔细想了想确实是三分，试着画画图发现它是一个凸性函数，存在一个最短距离。然后三分时间就可以了。

#include <stdio.h>#include <iostream>#include <map>#include <stack>#include <vector>#include <math.h>#include <string.h>#include <queue>#include <string>#include <stdlib.h>#include <algorithm>#define LL long long#define _LL __int64#define eps 1e-8#define PI acos(-1.0)using namespace std;const int maxn = 310;const int INF = 0x3f3f3f3f;struct node{double x,y,vx,vy;}p[maxn],tp[maxn];int n;double mindis,time;double dis(node a, node b){return sqrt((a.x - b.x)*(a.x - b.x) + (a.y - b.y)*(a.y - b.y));}double cal(double t){for(int i = 1; i <= n; i++){tp[i].x = p[i].x + t * p[i].vx;tp[i].y = p[i].y + t * p[i].vy;}double Max = -1.0;for(int i = 1; i < n; i++){for(int j = i+1; j <= n; j++){double ans = dis(tp[i], tp[j]);if(Max < ans)Max = ans;}}return Max;}void solve(){double low = 0, high = INF, mid,midmid;while(high - low > eps){mid = (high + low)/2;midmid = (mid + high)/2;double ans1 = cal(mid);double ans2 = cal(midmid);if(ans1 > ans2)low = mid;elsehigh = midmid;}time = low;mindis = cal(low);}int main(){int test;scanf("%d",&test);for(int item = 1; item <= test; item++){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);solve();printf("Case #%d: %.2lf %.2lf\n",item,time,mindis);}return 0;}

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