簡介：

在滿足約束條件$\varphi(x_1 ,x_2 ,\cdots ,x_n )=0$時求$f(x_1 ,x_2 ,\cdots ,x_n )$的極值。

結論：

令$L(x_1 ,x_2 ,\cdots ,x_n )=f(x_1 ,x_2 ,\cdots ,x_n )+\lambda \varphi(x_1 ,x_2 ,\cdots ,x_n )$。

再令$\frac{\partial L}{\partial {x_i}}$為L關於$x_i$的偏導數。（即視其他$x_j$為常量，僅對$x_i$求導）

則方程組

$\begin{cases}\frac{\partial L}{\partial {x_1}}=0\\\frac{\partial L}{\partial {x_2}}=0\\ \cdots \\ \frac{\partial L}{\partial {x_n}}=0 \\ \varphi(x_1 ,x_2 ,\cdots ,x_n )=0\end{cases}$

的某個解$(\lambda_{k},x_{k,1} ,x_{k,2} ,\cdots ,x_{k,n} )$一定對應著f的極值點。

實現：

先通過列舉/二分/三分確定$\lambda$，然後根據題意算出一組合法解更新答案即可。

程式碼（NOI2012騎行川藏）：

#include<bits/stdc++.h>
#define maxn 200005
#define maxm 500005
#define inf 0x7fffffff
#define eps 1e-14
#define ll long long
#define rint register int
#define debug(x) cerr<<#x<<": "<<x<<endl
#define
 
 fgx cerr<<"--------------"<<endl
#define dgx cerr<<"=============="<<endl

using namespace std;
struct road{double s,k,v;}A[maxn];
int n; double E;

inline int read(){
    int x=0,f=1; char c=getchar();
    for(;!isdigit(c);c=getchar()) if(c=='-') f=-1;
    for(;isdigit(c);c=getchar()) x=x*10
 
+c-'0';
    return x*f;
}

inline int dcmp(double x){return (abs(x)<=eps)?0:(x<0?-1:1);}
inline double phi(double lam){
    double res=0;
    for(int i=1;i<=n;i++){
        double l=max(A[i].v,0.0),r=1e5;
        while(dcmp(r-l)>0){
            double mid=(l+r)/2.0;
            if(dcmp(2.0*lam*A[i].k*(mid-A[i].v)*mid*mid-1.0)<=0) l=mid; else r=mid;
            //printf("%.8lf %.8lf %.8lf\n",2.0*lam*A[i].k*(mid-A[i].v)*mid*mid-1.0,l,r);
        }
        res+=A[i].k*(l-A[i].v)*(l-A[i].v)*A[i].s;
    }
    //printf("%lf\n",res-E);
    return res-E;
}
inline double calc(double lam){
    double res=0;
    for(int i=1;i<=n;i++){
        double l=max(A[i].v,0.0),r=1e5;
        while(dcmp(r-l)>0){
            double mid=(l+r)/2.0;
            if(dcmp(2*lam*A[i].k*(mid-A[i].v)*mid*mid-1)<=0) l=mid; else r=mid;
        }
        res+=A[i].s/l;
    }
    return res;
}


int main(){
    //freopen("bicycling17.in","r",stdin);
    n=read(),scanf("%lf",&E);
    //cout<<n<<":"<<E<<endl;
    for(int i=1;i<=n;i++)
        scanf("%lf%lf%lf",&A[i].s,&A[i].k,&A[i].v);
    double l=0,r=1e5;
    while(dcmp(r-l)>0){
        double mid=(l+r)/2.0;
        if(dcmp(phi(mid))>=0) l=mid; else r=mid;
    }
    //printf("%lf %lf\n",l,r);
    printf("%.6lf\n",calc(l));
    return 0;
}