[HDOJ3714]Error Curves（三分）

题目链接：http://acm.hdu.edu.cn/showproblem.php?pid=3714

题意：求n个二次函数在[0,1000]的最小值。

三分枚举。

 1 #include <bits/stdc++.h>
 2 using namespace std;
 3 
 4 const int maxn = 10100;
 5 const double eps = 1e-9;
 6 double a[maxn], b[maxn], c[maxn];
 7 int n;
 8 
 9 double f(double x) {
10   double ret = -1000000000;
11   for(int i = 1; i <= n; i++) {
12     double y = a[i]*x*x+b[i]*x+c[i];
13     ret = max(ret, y);
14   }
15   return ret;
16 }
17 
18 double ts() {
19   double lo = 0, hi = 1000.0;
20   double ret1, ret2;
21   while(hi - lo >= eps) {
22     double mid = (lo + hi) / 2.0;
23     double mimid = (mid + hi) / 2.0;
24     ret1 = f(mid);
25     ret2 = f(mimid);
26     if(ret1 > ret2) lo = mid;
27     else hi = mimid;
28   }
29   return ret1;
30 }
31 
32 int main() {
33   //freopen("in", "r", stdin);
34   int T;
35   scanf("%d", &T);
36   while(T--) {
37     scanf("%d", &n);
38     for(int i = 1; i <= n; i++) {
39       scanf("%lf %lf %lf", &a[i], &b[i], &c[i]);
40     }
41     printf("%.4lf\n", ts());
42   }
43   return 0;
44 }

 

posted @ 2016-10-13 20:34  Kirai  阅读(167)  评论(0)  编辑  收藏  举报