BZOJ 1492 货币兑换 Cash CDQ分治

这题n²算法就是一个维护上凸包的过程.

也可以用CDQ分治做.

我的CDQ分治做法和网上的不太一样,用左边的点建立一个凸包,右边的点在上面二分.

好处是思路清晰,避免了凸包的插入删除,坏处是多了一个log.

这题数据很水,同时注意精度.

#include<iostream>
#include<cstdio>
#include<cstdlib>
#include<cstring>
#include<cmath>
#include<ctime>
#include<string>
#include<iomanip>
#include<algorithm>
#include<map>
using namespace std;
#define LL long long
#define FILE "cash"
#define up(i,j,n) for(int i=j;i<=n;++i)
#define db double
#define ull unsigned long long
#define eps 1e-12
#define pii pair<int,int>
int read(){
	int x=0,f=1,ch=getchar();
	while(ch<'0'||ch>'9'){if(ch=='-')f=-1;ch=getchar();}
	while(ch>='0'&&ch<='9'){x=(x<<1)+(x<<3)+ch-'0';ch=getchar();}
	return f*x;
}
const int maxn=200010,maxm=20000,limit=1e6,mod=(int)(7+1e9+0.1);
const db inf=(1e18);
template<class T>bool cmax(T& a,T b){return a<b?a=b,true:false;}
template<class T>bool cmin(T& a,T b){return a>b?a=b,true:false;}
template<class T>T min(T& a,T& b){return a<b?a:b;}
template<class T>T max(T& a,T& b){return a>b?a:b;}
int n,N,M;
db f[maxn],X[maxn],Y[maxn],A[maxn],B[maxn],sp[maxn];
struct vec{
	db x,y;
	vec(db x=0,db y=0):x(x),y(y){}
	vec operator+(const vec& a){return vec(x+a.x,y+a.y);}
	vec operator-(const vec& a){return vec(x-a.x,y-a.y);}
	vec operator*(db b){return vec(x*b,y*b);}
}q[maxn],w[maxn];
int dcmp(db x){if(fabs(x)<eps)return 0;return x>0?1:-1;}
bool operator<(vec a,vec b){return a.x<b.x||(!dcmp(a.x-b.x)&&a.y<b.y);}
db dot(vec a,vec b){return a.x*b.x+a.y*b.y;}
db cro(vec a,vec b){return a.x*b.y-a.y*b.x;}
db len(vec a){return sqrt(dot(a,a));}
void build(int l,int r){
	N=0;
	up(i,l,r)q[++N]=vec(f[i]*X[i],f[i]*Y[i]);
	sort(q+1,q+N+1);
	M=0;
	w[++M]=q[1];
	up(i,2,N){
		while(M>1&&dcmp(cro(w[M]-w[M-1],q[i]-w[M]))>=0)M--;
		w[++M]=q[i];
	}
}
db getK(vec a,vec b){
	if(!dcmp(b.x-a.x))return inf;
	if(!dcmp(b.y-a.y))return -inf;
	return (b.y-a.y)/(b.x-a.x);
}
vec query(db sp){
	int left=2,right=M-1;
	if(getK(w[M],w[M-1])>sp)return w[M];
	if(getK(w[1],w[2])<sp)return w[1];
	while(left+1<right){
		int mid=(left+right)>>1;
		if(getK(w[mid],w[mid+1])<sp&&getK(w[mid-1],w[mid])>sp)return w[mid];
		if(getK(w[mid],w[mid-1])<sp&&getK(w[mid],w[mid+1])<sp)right=mid;
		else left=mid;
	}
	int mid=left;
	if(getK(w[mid],w[mid+1])<sp&&getK(w[mid-1],w[mid])>sp)return w[mid];
	mid=right;
	if(getK(w[mid],w[mid+1])<sp&&getK(w[mid-1],w[mid])>sp)return w[mid];
	return w[left];
}
void cdq(int l,int r){
	if(l==r){
		cmax(f[l],f[l-1]);
		//printf("%d %.3lf\n",l,f[l]);
		return;
	}
	int mid=(l+r)>>1;
	cdq(l,mid);
	build(l,mid);
	up(i,mid+1,r){
		vec k=query(sp[i]);
		cmax(f[i],f[i-1]);
		cmax(f[i],k.x*A[i]+k.y*B[i]);
	}
	cdq(mid+1,r);
}
int main(){
	freopen(FILE".in","r",stdin);
	freopen(FILE".out","w",stdout);
	scanf("%d%lf",&n,&f[0]);
	up(i,1,n){
		db x,y,k;
		scanf("%lf%lf%lf",&x,&y,&k);
		A[i]=x,B[i]=y;
		X[i]=k/(x*k+y);
		Y[i]=1.0/(x*k+y);
		sp[i]=-A[i]/B[i];
	}
	cdq(1,n);
	printf("%.3lf\n",f[n]);
	return 0;
}