HDU 4717 The Moving Points

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  ^thpoint, 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

求的是所有点之间的最大距离最小，考虑对于任意两个点来说，

它们之间的距离可以通过三分时间来确定最小值。

那么所有点之间的最大值也会呈现抛物线的形状。

通过三分时间，就可以直接求出距离了。

#include<map> #include<set> #include<ctime> #include<cmath> #include<stack> #include<queue> #include<string> #include<vector> #include<cstdio> #include<cstring> #include<iostream> #include<algorithm> #include<functional> using namespace std; #define ms(x,y) memset(x,y,sizeof(x)) #define rep(i,j,k) for(int i=j;i<=k;i++) #define per(i,j,k) for(int i=j;i>=k;i--) #define loop(i,j,k) for (int i=j;i!=-1;i=k[i]) #define inone(x) scanf("%d",&x) #define intwo(x,y) scanf("%d%d",&x,&y) #define inthr(x,y,z) scanf("%d%d%d",&x,&y,&z) #define infou(x,y,z,p) scanf("%d%d%d%d",&x,&y,&z,&p) #define lson x<<1,l,mid #define rson x<<1|1,mid+1,r #define mp(i,j) make_pair(i,j) #define ff first #define ss second typedef long long LL; typedef pair<int, int> pii; const int low(int x) { return x&-x; } const int INF = 0x7FFFFFFF; const int mod = 1e9 + 7; const int N = 1e5 + 10; const double eps = 1e-8; int T, n, cas = 1; int x[N], y[N], vx[N], vy[N]; double get(int i, double a, int j) { double xi = x[i] + a*vx[i], yi = y[i] + a*vy[i]; double xj = x[j] + a*vx[j], yj = y[j] + a*vy[j]; return sqrt((xi - xj)*(xi - xj) + (yi - yj)*(yi - yj)); } int main() { for (inone(T); T--; cas++) { inone(n); rep(i, 1, n) infou(x[i], y[i], vx[i], vy[i]); double l = 0, r = 1e6, d = INF; while (r - l > eps) { double a = (l + l + r) / 3, b = (l + r + r) / 3; double da = 0, db = 0; rep(i, 1, n) rep(j, i + 1, n) { double d1 = get(i, a, j), d2 = get(i, b, j); da = max(da, d1); db = max(db, d2); } if (da > db) l = a; else r = b; d = min(da, db); } printf("Case #%d: %.2lf %.2lf\n",cas, r, d); } return 0; }