hdu4498 Function Curve 分段simpson

题目：给出k1,k2,…,kn, a1,a2,…,an 和 b1,b2,…,bn 
求F(x)=min{100,min{ki*(x-ai)^2+bi | 0 < i <= n}} 在坐标上画出的曲线的长度。

要求x>=0&&x<=100，n<=50

思路：求出所有函数之间的交点，所有函数与y=100的交点，交点排序，累加相邻交点的曲线长度

代码：

#pragma comment(linker, "/STACK:1024000000,1024000000")
#include<iostream>
#include<algorithm>
#include<ctime>
#include<cstdio>
#include<cmath>
#include<cstring>
#include<string>
#include<vector>
#include<map>
#include<set>
#include<queue>
#include<stack>
#include<list>
#include<numeric>
using namespace std;
#define LL long long
#define ULL unsigned long long
#define INF 0x3f3f3f3f
#define mm(a,b) memset(a,b,sizeof(a))
#define PP puts("*********************");
template<class T> T f_abs(T a){ return a > 0 ? a : -a; }
template<class T> T gcd(T a, T b){ return b ? gcd(b, a%b) : a; }
template<class T> T lcm(T a,T b){return a/gcd(a,b)*b;}
// 0x3f3f3f3f3f3f3f3f
// 0x3f3f3f3f

const double eps=1e-5;//根据实际情况修改
const int maxn=105;
double k[maxn],a[maxn],b[maxn];
vector<double> inter;
int n;
double sqr(double x){
    return x*x;
}
int dblcmp(double x){//判断正负0
    if(f_abs(x)<eps) return 0;
    else return x<0?-1:1;
}
void judge(double x){
    if(dblcmp(x)>=0&&dblcmp(x-100)<=0)
        inter.push_back(x);
}
void getinter(){
    inter.clear();
    inter.push_back(0);
    inter.push_back(100);
    for(int i=0;i<n;i++){
        if(b[i]>100) continue;
        double x1=a[i]+sqrt((100-b[i])/k[i]);
        double x2=a[i]-sqrt((100-b[i])/k[i]);
        judge(x1);
        judge(x2);
    }
    for(int i=0;i<n;i++)
        for(int j=i+1;j<n;j++){
            double A=k[i]-k[j];
            double B=2*k[j]*a[j]-2*k[i]*a[i];
            double C=k[i]*a[i]*a[i]+b[i]-k[j]*a[j]*a[j]-b[j];
            if(dblcmp(A)==0){
                if(dblcmp(B)!=0)
                    judge(-C/B);
                continue;
            }
            else{
                double delta=B*B-4*A*C;
                if(dblcmp(delta)<0) continue;
                if(dblcmp(delta)==0) judge(-B/(2*A));
                else{
                    delta=sqrt(delta);
                    double x1=(-B+delta)/(2*A);
                    double x2=(-B-delta)/(2*A);
                    judge(x1);
                    judge(x2);
                }
            }
        }
}
int pos;
double F(double x){//
	//Simpson公式用到的函数
    return sqrt(1+sqr(2*k[pos]*(x-a[pos])));
}
double simpson(double a, double b){
	double c = a + (b - a) / 2;
	return (F(a) + 4 * F(c) + F(b))*(b - a) / 6;
}
double asr(double a, double b, double eps, double A){
	double c = a + (b - a) / 2;
	double L = simpson(a, c), R = simpson(c, b);
	if (fabs(L + R - A) <= 15 * eps) return L + R + (L + R - A) / 15.0;
	return asr(a, c, eps / 2, L) + asr(c, b, eps / 2, R);
}
double asr(double a, double b, double eps){
	return asr(a, b, eps, simpson(a, b));
}
double Function(double x,int i){
    return k[i]*sqr(x-a[i])+b[i];
}
int main(){

    int T;
    scanf("%d",&T);
    while(T--){
        scanf("%d",&n);
        for(int i=0;i<n;i++)
            scanf("%lf%lf%lf",&k[i],&a[i],&b[i]);
        getinter();
        sort(inter.begin(),inter.end());
        int sz=inter.size();
        double ans=0;
        for(int i=1;i<sz;i++){
            double x1=inter[i-1],x2=inter[i];
            if(dblcmp(x1-x2)==0) continue;
            double x=(x1+x2)/2;
            pos=0;
            for(int j=1;j<n;j++){
                if(Function(x,j)<Function(x,pos))
                    pos=j;
            }
            if(dblcmp(Function(x,pos)-100)<0)
                ans+=asr(x1,x2,eps);
            else
                ans+=x2-x1;
        }
        printf("%.2f\n",ans);
    }
    return 0;
}