UVaLive 5009 Error Curves (三分)

题意：给定 n 条二次曲线， fi(x) = aix^2 + bix + c 定义 F(x) =max{Si(x)}，求 F(x) 在 0 ~ 1000 上的最小值。

析：从题目给定的曲线上进行分析，很容易知道，最后的所形成的图形一定是下凸的，而这个图形就一定有一个最小值，而下凸函数可以用三分来求解。

代码如下：

#pragma comment(linker, "/STACK:1024000000,1024000000")
#include <cstdio>
#include <string>
#include <cstdlib>
#include <cmath>
#include <iostream>
#include <cstring>
#include <set>
#include <queue>
#include <algorithm>
#include <vector>
#include <map>
#include <cctype>
#include <cmath>
#include <stack>
#include <sstream>
#include <list>
#include <assert.h>
#include <bitset>
#include <numeric>
#define debug() puts("++++")
#define gcd(a, b) __gcd(a, b)
#define lson l,m,rt<<1
#define rson m+1,r,rt<<1|1
#define fi first
#define se second
#define pb push_back
#define sqr(x) ((x)*(x))
#define ms(a,b) memset(a, b, sizeof a)
#define sz size()
#define pu push_up
#define pd push_down
#define cl clear()
#define lowbit(x) -x&x
//#define all 1,n,1
#define FOR(i,n,x)  for(int i = (x); i < (n); ++i)
#define freopenr freopen("in.in", "r", stdin)
#define freopenw freopen("out.out", "w", stdout)
using namespace std;

typedef long long LL;
typedef unsigned long long ULL;
typedef pair<int, int> P;
const int INF = 0x3f3f3f3f;
const LL LNF = 1e17;
const double inf = 1e20;
const double PI = acos(-1.0);
const double eps = 1e-10;
const int maxn = 10000 + 5;
const int maxm = 1e6 + 2;
const LL mod = 1000000007;
const int dr[] = {-1, 1, 0, 0, 1, 1, -1, -1};
const int dc[] = {0, 0, 1, -1, 1, -1, 1, -1};
const char *de[] = {"0000", "0001", "0010", "0011", "0100", "0101", "0110", "0111", "1000", "1001", "1010", "1011", "1100", "1101", "1110", "1111"};
int n, m;
const int mon[] = {0, 31, 28, 31, 30, 31, 30, 31, 31, 30, 31, 30, 31};
const int monn[] = {0, 31, 29, 31, 30, 31, 30, 31, 31, 30, 31, 30, 31};
inline bool is_in(int r, int c) {
  return r >= 0 && r < n && c >= 0 && c < m;
}

int a[maxn], b[maxn], c[maxn];

double F(double x){
  double ans = a[0] * sqr(x) + b[0] * x + c[0];
  for(int i = 1; i < n; ++i)
    ans = max(ans, a[i] * sqr(x) + b[i] * x + c[i]);
  return ans;
}

int main(){
  int T;  cin >> T;
  while(T--){
    scanf("%d", &n);
    for(int i = 0; i < n; ++i)  scanf("%d %d %d", a+i, b+i, c+i);
    double l = 0., r = 1000.;
    while(r - l > eps){
      double m1 = l + (r-l)/3.;
      double m2 = r - (r-l)/3;
      if(F(m1) < F(m2))  r = m2;
      else l = m1;
    }
    printf("%.4f\n", F(l));
  }
  return 0;
}