【知识点】拉格朗日乘数法
简介:
在满足约束条件$\varphi(x_1 ,x_2 ,\cdots ,x_n )=0$时求$f(x_1 ,x_2 ,\cdots ,x_n )$的极值。
结论:
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$\frac{\partial L}{\partial {x_i}}$为L关于$x_i$的偏导数。(即视其他$x_j$为常量,仅对$x_i$求导)
则方程组
$\begin{cases}\frac{\partial L}{\partial {x_1}}=0\\\frac{\partial L}{\partial {x_2}}=0\\ \cdots \\ \frac{\partial L}{\partial {x_n}}=0 \\ \varphi(x_1 ,x_2 ,\cdots ,x_n )=0\end{cases}$
的某个解$(\lambda_{k},x_{k,1} ,x_{k,2} ,\cdots ,x_{k,n} )$一定对应着f的极值点。
实现:
先通过枚举/二分/三分确定$\lambda$,然后根据题意算出一组合法解更新答案即可。
代码(NOI2012骑行川藏):
![](https://images.cnblogs.com/OutliningIndicators/ContractedBlock.gif)
#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; }