hdu 4717 The Moving Points(三分)

题目连接：hdu 4717 The Moving Points

题目大意：给出n个点，每一个点有初始的位置(x,y),以及单位时间内移动的距离，向量形式给出。且在哪一个时刻中，n个点之间两两距离的最大值最小，最小值为多少。

解题思路：类似与二分算法的三分，由于假设将时间t和所要求的两两之间距离的最大值d做成一个函数曲线，单调性应该是先递减后递增的，所以用三分法求极值。

#include <cstdio>
#include <cstring>
#include <cmath>
#include <algorithm>

using namespace std;
const int N = 305;
const double eps = 1e-6;
struct point {
double x;
double y;
}s
, p
;

int n;

void init () {
scanf("%d", &n);
for (int i = 0; i < n; i++)
scanf("%lf%lf%lf%lf", &s[i].x, &s[i].y, &p[i].x, &p[i].y);
}

inline double dis (double x, double y) {
return sqrt(x*x+y*y);
}

double cat (double k) {
double ans = 0;
for (int i = 0; i < n; i++) {
double xi = s[i].x + p[i].x * k;
double yi = s[i].y + p[i].y * k;

for (int j = i + 1; j < n; j++) {
double pi = s[j].x + p[j].x * k;
double qi = s[j].y + p[j].y * k;
ans = max(ans, dis(xi-pi, yi-qi));
}
}
return ans;
}

void solve () {
double l = 0;
double r = 0xffffff;

while (fabs(r-l) > eps) {
double tmp = (r-l)/3;
double midl = l + tmp;
double midr = r - tmp;
if (cat(midl) < cat(midr))
r = midr;
else
l = midl;
}
printf("%.2lf %.2lf\n", l, cat(l));
}

int main () {
int cas;
scanf("%d", &cas);
for (int i = 1; i <= cas; i++) {
init ();
printf("Case #%d: ", i);
solve();
}
return 0;
}