给(1 ≤ n ≤ 100 000 )个向量 (ai,bi)和向量(p,q),现在k1a1+k2a2+...+knan≥p且k1b1+k2b2+...+knbn≥p,求min(∑iki)
我们考虑把这n个向量画出来,再添上(0,maxy)和(maxx,0),求个凸包
转自harryguo
现在有一个结论,
1.到凸包内任意一点,系数和<1
2.到凸包上任意一点,系数和=1
于是可以二分凸包倍增比,或者直接连线求交点(我是这么做的)。
#include<bits/stdc++.h>
using namespace std;
#define For(i,n) for(int i=1;i<=n;i++)
#define Fork(i,k,n) for(int i=k;i<=n;i++)
#define Rep(i,n) for(int i=0;i<n;i++)
#define ForD(i,n) for(int i=n;i;i--)
#define ForkD(i,k,n) for(int i=n;i>=k;i--)
#define RepD(i,n) for(int i=n;i>=0;i--)
#define Forp(x) for(int p=Pre[x];p;p=Next[p])
#define Forpiter(x) for(int &p=iter[x];p;p=Next[p])
#define Lson (o<<1)
#define Rson ((o<<1)+1)
#define MEM(a) memset(a,0,sizeof(a));
#define MEMI(a) memset(a,127,sizeof(a));
#define MEMi(a) memset(a,128,sizeof(a));
#define INF (2139062143)
#define F (100000007)
#define pb push_back
#define mp make_pair
#define fi first
#define se second
#define vi vector<int>
#define pi pair<int,int>
#define SI(a) ((a).size())
typedef long long ll;
typedef unsigned long long ull;
ll mul(ll a,ll b){return (a*b)%F;}
ll add(ll a,ll b){return (a+b)%F;}
ll sub(ll a,ll b){return (a-b+llabs(a-b)/F*F+F)%F;}
void upd(ll &a,ll b){a=(a%F+b%F)%F;}
int read()
{
int x=0,f=1; char ch=getchar();
while(!isdigit(ch)) {if (ch=='-') f=-1; ch=getchar();}
while(isdigit(ch)) { x=x*10+ch-'0'; ch=getchar();}
return x*f;
}
int n,p,q;
map<ll ,ll > h;
map<ll ,ll >::iterator it;
struct point{
ll x,y;
point(pair<ll,ll> p):x(p.fi),y(p.se) { }
friend ll operator*(point a,point b){
return a.x*b.y-a.y*b.x;
}
};
#define EPS (1E-10)
#define MAXN (100000+10)
#define cross(p1,p2,p3) (((p2).x-(p1).x)*((p3).y-(p2).y)-((p3).x-(p2).x)*((p2).y-(p1).y))
#define crossOp(p1,p2,p3) sign(cross(p1,p2,p3))
inline int sign(double a) {
return a<-EPS?-1:a>EPS;
}
pair<ll,ll> st[MAXN];
int sz=0;
int main()
{
// freopen("CF605C.in","r",stdin);
// freopen(".out","w",stdout);
ll ans;
n=read(),p=read(),q=read();
ll maxx=0,maxy=0;
For(i,n) {
ll a=read(),b=read();
maxx=max(maxx,a);
maxy=max(maxy,b);
if (h.count(a)) h[a]=max(h[a],b);
else h[a]=b;
ans=max(p/a,q/b);
}
st[++sz]=mp(0,maxy);
for(it=h.begin();it!=h.end();it++) {
pair<ll,ll> pa=*it;
while(sz>=2&&crossOp( ((point)st[sz-1]), ((point)st[sz]), ((point)pa) )>=0LL )
{
--sz;
}st[++sz]=pa;
}
{
pair<ll,ll> pa=mp(maxx,0);
while(sz>=2&&crossOp(((point)st[sz-1]), ((point)st[sz]), ((point)pa) )>=0) --sz;
st[++sz]=pa;
}
point pa(mp(p,q)),O(mp(0,0));
for(int i=2;i<=sz;i++) {
if ((crossOp( pa , O, ((point)st[i]) )) * (crossOp( pa , O, ((point)st[i-1]) )) <=0 ) {
double t=(point)st[i]*(point)st[i-1];
t/=q*(st[i].fi-st[i-1].fi)+p*(st[i-1].se-st[i].se);
cout<<setprecision(10)<<setiosflags(ios::fixed);
cout<<1/t<<endl;
return 0;
}
}
return 0;
}