BZOJ1502: [NOI2005]月下柠檬树

自适应辛普森积分

总体思想就是先用一个二次函数拟合曲线然后利用二次函数积分来计算
然后就一直细分下去直到左右两块分开积分 与整块积分的误差小于eps就可以了

#include<iostream>
#include<cstdio>
#include<cstdlib>
#include<cstring>
#include<algorithm>
#include<cmath>
using namespace std;
const
 int MAXN=550;
const
 double eps=1e-6;
int n; 
struct Point
{double x,y;}P[MAXN];

struct Circle
{Point o;double r;}C[MAXN];

struct Line
{
    Point x,y;
    double k,b;
    inline void Bg()
    {
        k=(y.y-x.y)/(y.x-x.x);
        b=y.y-k*y.x;
    }
    inline double D(double x)
    {return k*x+b;}

}L[MAXN];
int Con=0;
#define abs(a) ((a)<0?(-(a)):(a))
inline double D(double x)
{
    double d,ans=0;
    for(int i=1;i<=n;i++)
    {
        d=abs(x-C[i].o.x);
        if(d-C[i].r>-eps)
            continue;
        ans=max(ans,2*sqrt(C[i].r*C[i].r-d*d));
    }
    for(int i=1;i<=Con;i++)
        if(x>=L[i].x.x&&x<=L[i].y.x)
            ans=max(ans,2*L[i].D(x));
    return ans;
}
#define Calc(a,b,c,d) (((b)+4*(c)+(d))*(a)/6)
double Simpson(double L,double M,double R,double fL,double fM,double fR,double sqr)
{
    double Mid1=(L+M)/2,Mid2=(M+R)/2,D1=D(Mid1),D2=D(Mid2),Area1=Calc(M-L,fL,D1,fM),Area2=Calc(R-M,fM,D2,fR);
    if(abs(Area1+Area2-sqr)<eps) return Area1+Area2;
    return Simpson(L,Mid1,M,fL,D1,fM,Area1)+Simpson(M,Mid2,R,fM,D2,fR,Area2);
}

int main()
{
    double Min=1<<29,Max=-Min,d,h,Ang,Sin,Cos;
    scanf("%d%lf",&n,&Ang);
    n++;
    for(int i=1;i<=n;i++) 
        scanf("%lf",&h),C[i].o.y=0,C[i].o.x=h/tan(Ang)+C[i-1].o.x;
    for(int i=1;i<n;i++)scanf("%lf",&C[i].r);
    for(int i=1;i<=n;i++)
        Min=min(Min,C[i].o.x-C[i].r),Max=max(Max,C[i].o.x+C[i].r);
    for(int i=1;i<n;i++)
    {
        d=C[i+1].o.x-C[i].o.x;
        if(d-abs(C[i+1].r-C[i].r)<-eps)continue;
        Sin=(C[i].r-C[i+1].r)/d,Cos=sqrt(1-Sin*Sin);
        L[++Con]=(Line){(Point){C[i].o.x+C[i].r*Sin,C[i].r*Cos},(Point){C[i+1].o.x+C[i+1].r*Sin,C[i+1].r*Cos}};
        L[Con].Bg();
    }
    double L=Min,R=Max;
    double M=(L+R)/2;
    double fL=D(L),fM=D(M),fR=D(R);
    printf("%.2lf\n",Simpson(L,M,R,fL,fM,fR,Calc(R-L,fL,fM,fR)));
    return 0;
}