hdu 4498 自适应simpson

传送门

题意： 
给出k1,k2,…,kn, a1,a2,…,an 和 b1,b2,…,bn 
求函数： 
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。

限制： 
1 <= n <= 50; 0 <= ai,bi < 100; 0 < ki < 100

思路： 
先求出所有交点，然后排序，把函数分段，然后利用 
L=∫[a，b] √[1+(f ‘(x))^2] dx 
利用自适应simpson进行长度积分。

//china no.1
#pragma comment(linker, "/STACK:1024000000,1024000000")
#include <vector>
#include <iostream>
#include <string>
#include <map>
#include <stack>
#include <cstring>
#include <queue>
#include <list>
#include <stdio.h>
#include <set>
#include <algorithm>
#include <cstdlib>
#include <cmath>
#include <iomanip>
#include <cctype>
#include <sstream>
#include <functional>
#include <stdlib.h>
#include <time.h>
#include <bitset>
using namespace std;

#define pi acos(-1)
#define s_1(x) scanf("%d",&x)
#define s_2(x,y) scanf("%d%d",&x,&y)
#define s_3(x,y,z) scanf("%d%d%d",&x,&y,&z)
#define s_4(x,y,z,X) scanf("%d%d%d%d",&x,&y,&z,&X)
#define S_1(x) scan_d(x)
#define S_2(x,y) scan_d(x),scan_d(y)
#define S_3(x,y,z) scan_d(x),scan_d(y),scan_d(z)
#define PI acos(-1)
#define endl '\n'
#define srand() srand(time(0));
#define me(x,y) memset(x,y,sizeof(x));
#define foreach(it,a) for(__typeof((a).begin()) it=(a).begin();it!=(a).end();it++)
#define close() ios::sync_with_stdio(0); cin.tie(0);
#define FOR(x,n,i) for(int i=x;i<=n;i++)
#define FOr(x,n,i) for(int i=x;i<n;i++)
#define fOR(n,x,i) for(int i=n;i>=x;i--)
#define fOr(n,x,i) for(int i=n;i>x;i--)
#define W while
#define sgn(x) ((x) < 0 ? -1 : (x) > 0)
#define bug printf("***********\n");
#define db double
#define ll long long
#define mp make_pair
#define pb push_back
typedef long long LL;
typedef pair <int, int> ii;
const int INF=0x3f3f3f3f;
const LL LINF=0x3f3f3f3f3f3f3f3fLL;
const int dx[]={-1,0,1,0,1,-1,-1,1};
const int dy[]={0,1,0,-1,-1,1,-1,1};
const int maxn=4e3+10;
const int maxx=4e5+10;
const double EPS=1e-8;
const double eps=1e-8;
const int mod=1e9+7;
template<class T>inline T min(T a,T b,T c) { return min(min(a,b),c);}
template<class T>inline T max(T a,T b,T c) { return max(max(a,b),c);}
template<class T>inline T min(T a,T b,T c,T d) { return min(min(a,b),min(c,d));}
template<class T>inline T max(T a,T b,T c,T d) { return max(max(a,b),max(c,d));}
template <class T>
inline bool scan_d(T &ret){char c;int sgn;if (c = getchar(), c == EOF){return 0;}
while (c != '-' && (c < '0' || c > '9')){c = getchar();}sgn = (c == '-') ? -1 : 1;ret = (c == '-') ? 0 : (c - '0');
while (c = getchar(), c >= '0' && c <= '9'){ret = ret * 10 + (c - '0');}ret *= sgn;return 1;}

inline bool scan_lf(double &num){char in;double Dec=0.1;bool IsN=false,IsD=false;in=getchar();if(in==EOF) return false;
while(in!='-'&&in!='.'&&(in<'0'||in>'9'))in=getchar();if(in=='-'){IsN=true;num=0;}else if(in=='.'){IsD=true;num=0;}
else num=in-'0';if(!IsD){while(in=getchar(),in>='0'&&in<='9'){num*=10;num+=in-'0';}}
if(in!='.'){if(IsN) num=-num;return true;}else{while(in=getchar(),in>='0'&&in<='9'){num+=Dec*(in-'0');Dec*=0.1;}}
if(IsN) num=-num;return true;}

void Out(LL a){if(a < 0) { putchar('-'); a = -a; }if(a >= 10) Out(a / 10);putchar(a % 10 + '0');}
void print(LL a){ Out(a),puts("");}
//freopen( "in.txt" , "r" , stdin );
//freopen( "data.txt" , "w" , stdout );
//cerr << "run time is " << clock() << endl;


int pos,n;
db A[105],B[105],C[105];

vector<db>Q;

void cal(double a,double b,double c)
{	//计算a*x^2+b*x+c=0的解
	if(a==0 && b==0)
		return ;
	if(a==0)
    {
		double t=-c/b;
		if(t>=0 && t<=100)
			Q.pb(t);
		return ;
	}
	double delta=b*b-4*a*c;
	if(delta<0) return ;
	if(delta==0)
	{
		double t=-b/(2*a);
		if(t>=0 && t<=100)
			Q.pb(t);
	}
	else
	{
		double t1=(-b+sqrt(delta))/(2*a);
		double t2=(-b-sqrt(delta))/(2*a);
		if(t1>=0 && t1<=100)
			Q.pb(t1);
		if(t2>=0 && t2<=100)
			Q.pb(t2);
	}
}

int get_pos(db x)
{
    int p=0;
    db minn=100;
    FOR(1,n,i)
    {
        db tmp=A[i]*x*x+B[i]*x+C[i];
        if(tmp<minn)
        {
            minn=tmp;
            p=i;
        }
    }
    return p;
}

double F(double x1)
{
    return sqrt(1.0+(x1*2*A[pos]+B[pos])*(x1*2*A[pos]+B[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);
    double R = simpson(c,b);
    if(fabs(L+R-A) <= 15*eps)return L+R+(L+R-A)/15;
    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));
}

int t;
db a,b,k;
int main()
{
    s_1(t);
    W(t--)
    {
        s_1(n);
        A[0]=0;
        B[0]=0;
        C[0]=100;
        FOR(1,n,i)
        {
            scanf("%lf%lf%lf",&k,&a,&b);
            A[i]=k;
            B[i]=-2.0*a*k;
            C[i]=k*a*a+b;
        }
        Q.clear();
        FOr(0,n,i)
        {
            FOR(i+1,n,j)
            {
                cal(A[i]-A[j],B[i]-B[j],C[i]-C[j]);
            }
        }
        db ans=0;
        Q.pb(0);
        Q.pb(100);
        sort(Q.begin(),Q.end());
        FOr(0,Q.size()-1,i)
        {
            db l=Q[i];db r=Q[i+1];
            if(fabs(r-l)<eps) continue;
            db mid=(l+r)/2;
            pos=get_pos(mid);
            //cout<<pos<<endl;
            ans+=asr(l,r,eps);
            //printf("%f\n",ans);
        }
        printf("%.2f\n",ans);
    }
}