HDU - 4498 Function Curve 自适应simpson

题目链接：点击打开链接

题意：求题目中F(x)的弧长

思路：求出各个二次函数的交点并保存起来，然进行排序，然后按各个交点求出以这个点为端点的一段最小的函数，并求出这个函数在这个区间的弧长。弧长当然可以用自适应simpson模版

代码：

#include <iostream>
#include <cstdio>
#include <cmath>
#include <algorithm>
#include <vector>
using namespace std;
const int maxn=1e3;
int bext;
double k[maxn],a[maxn],b[maxn];
double han(double x){
return 2*k[bext]*x+a[bext];
}
double F(double x){
return sqrt(1.0+han(x)*han(x));
}
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));
}
vector<double> x;
void push(double t){
if (t>0.0&&t<100.0)
x.push_back(t);
}
void jfc(double a,double b,double c){
if(fabs(a)<1e-8&&fabs(b)<1e-8)
return;
double t=b*b-4.0*a*c;
if (t<-1e-8)
return ;
if(fabs(a)<1e-8)
push(-c/b);
else{
push((-b-sqrt(t))/2/a);
if (t>1e-8)
push((-b+sqrt(t))/2/a);
}
}
int main(int argc, const char * argv[]) {
int t,n;
cin>>t;
while (t--) {
cin>>n;
double ans=0;
for (int i=1; i<=n; ++i){
cin>>k[i]>>a[i]>>b[i];
b[i]=k[i]*a[i]*a[i]+b[i];
a[i]=-2*a[i]*k[i];
}
x.clear();
x.push_back(0.0);
x.push_back(100.0);
k[0]=a[0]=0,b[0]=100.0;
for (int i=0; i<n; ++i)
for (int j=i+1; j<=n; ++j)
jfc(k[j]-k[i],a[j]-a[i],b[j]-b[i]);
int s=x.size();
sort(x.begin(), x.end());
// for (int i=0; i<s; ++i) {
// cout<<x[i]<<" ";
//}
// cout<<endl;
double xie=101;
for (int i=0; i<s-1; ++i) {
double t=101;
for (int j=0;j<=n; ++j) {
double tt=k[j]*x[i]*x[i]+a[j]*x[i]+b[j];
if (fabs(t-tt)<=1e-8||t>tt) {
double xx=2*k[j]*x[i]+a[j];
if (fabs(t-tt)<1e-8&&(xie+1e-8)<=xx)
continue;
bext=j;
t=tt;
xie=xx;
}
}
//cout<<bext<<endl;
ans+=asr(x[i],x[i+1],1e-8);
}
printf("%.2f\n",ans);
}
return 0;
}