LA5009 Error Curves

Posted 2020-10-22 嘒彼小星

题目大意：给定n个二次函数fi(x)，规定F(x) = max(fi(x))，求F在[0,1000]上的max

 

其实F（x）也是一个单峰函数，因为如果出现双峰，显然有一个可以继续往上走把另一个峰遮住，即另一个峰不可能是max

三分即可

被精度卡了半天

 1 #include <iostream>
 2 #include <cstdio>
 3 #include <cstring>
 4 #include <cstdlib>
 5 #include <algorithm>
 6 #include <queue>
 7 #include <vector>
 8 #include <map>
 9 #include <string> 
10 #include <cmath> 
11 #define min(a, b) ((a) < (b) ? (a) : (b))
12 #define max(a, b) ((a) > (b) ? (a) : (b))
13 #define abs(a) ((a) < 0 ? (-1 * (a)) : (a))
14 template<class T>
15 inline void swap(T &a, T &b)
16 {
17     T tmp = a;a = b;b = tmp;
18 }
19 inline void read(int &x)
20 {
21     x = 0;char ch = getchar(), c = ch;
22     while(ch < ‘0‘ || ch > ‘9‘) c = ch, ch = getchar();
23     while(ch <= ‘9‘ && ch >= ‘0‘) x = x * 10 + ch - ‘0‘, ch = getchar();
24     if(c == ‘-‘) x = -x;
25 }
26 
27 const double INF = 0x3f3f3f3f;
28 const double eps = 0.0000000001;
29 
30 
31 int t,n,a[10100],b[10100],c[10010];
32 
33 double f(double x)
34 {
35     double ma = x * x * a[1] + x * b[1] + c[1], t;
36     for(int i = 2;i <= n;++ i) 
37     {
38         t = x * x * a[i] + x * b[i] + c[i];
39         //if(!cmp(t, ma)) ma = t;
40         ma = max(ma, t);
41     }
42     return ma;
43 }
44 
45 int main()
46 {
47     read(t);
48     for(;t;--t)
49     {
50         read(n);
51         for(int i = 1;i <= n;++ i) read(a[i]), read(b[i]), read(c[i]);
52         double l = 0, r = 1000, ans, mid1, mid2;
53         while(r - l - eps >= 0)
54         {
55             mid1 = l + (r - l) / 3.0;mid2 = r - (r - l) / 3.0;
56             if(f(mid1) - f(mid2) >= 0) l = mid1;
57             else r = mid2;
58         }
59         printf("%.4lf\n", f(l));
60     }
61     return 0;
62 }

LA5009