【BZOJ】【1492】【NOI207】货币兑换Cash

DP/CDQ分治

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　　orz Hzwer

　　copy了下他的代码……结果在while(j<top......)这一句中把一个括号的位置打错了……找了我一个多小时才找到TAT

　　很神奇……顺便贴下CDQ的论文吧

 /**************************************************************
     Problem: 1492
     User: Tunix
     Language: C++
     Result: Accepted
     Time:1420 ms
     Memory:13388 kb
 ****************************************************************/

 //BZOJ 1492
 #include<cmath>
 #include<cstdio>
 #include<cstdlib>
 #include<iostream>
 #include<algorithm>
 #define rep(i,n) for(int i=0;i<n;++i)
 #define F(i,j,n) for(int i=j;i<=n;++i)
 #define D(i,j,n) for(int i=j;i>=n;--i)
 using namespace std;
 inline int getint(){
     int v=,sign=; char ch=getchar();
     while(ch<''||ch>''){ if (ch=='-') sign=-; ch=getchar();}
     while(ch>=''&&ch<=''){ v=v*+ch-''; ch=getchar();}
     return v*sign;
 }
 const int N=1e5+,INF=~0u>>;
 const double eps=1e-;
 typedef long long LL;
 /******************tamplate*********************/
 int n,top,st[N];
 double f[N];
 struct Point{
     double x,y,a,b,k,rate;
     int w,id;
     void out(){
         printf("%lf %lf %lf %lf %lf %lf\n",x,y,a,b,k,rate);
     }
 }p[N],t[N];
 double getk(int a,int b){
     if (!b)return -1e20;
     if (fabs(p[a].x-p[b].x)<eps)return 1e20;
     return (p[b].y-p[a].y)/(p[b].x-p[a].x);
 }
 inline bool operator < (Point a,Point b){return a.k>b.k;}
 void solve(int l,int r){
     if (l==r){
         f[l]=max(f[l],f[l-]);
         p[l].y=f[l]/(p[l].a*p[l].rate+p[l].b);
         p[l].x=p[l].rate*p[l].y;
         return;
     }
     int l1,l2,mid=(l+r)>>,j=;
     l1=l; l2=mid+;
     F(i,l,r)
         if (p[i].id<=mid) t[l1++]=p[i];
         else t[l2++]=p[i];
     F(i,l,r) p[i]=t[i];

     solve(l,mid);
     top=;
     F(i,l,mid){
         while(top>&&getk(st[top-],st[top])<getk(st[top-],i)+eps)
             top--;
         st[++top]=i;
     }
     st[++top]=;
     F(i,mid+,r){
         while(j<top && getk(st[j],st[j+])+eps>p[i].k)j++;
         f[p[i].id]=max(f[p[i].id],
                 p[st[j]].x*p[i].a+p[st[j]].y*p[i].b);
     }
     solve(mid+,r);
     l1=l;l2=mid+;
     F(i,l,r)
         if(((p[l1].x<p[l2].x||(fabs(p[l1].x-p[l2].x)<eps&&p[l1].y<p[l2].y))||l2>r)&&l1<=mid)t[i]=p[l1++];
         else t[i]=p[l2++];
     F(i,l,r) p[i]=t[i];
 }

 int main(){
 #ifndef ONLINE_JUDGE
     freopen("1492.in","r",stdin);
     freopen("1492.out","w",stdout);
 #endif
     scanf("%d%lf",&n,&f[]);
     F(i,,n){
         scanf("%lf%lf%lf",&p[i].a,&p[i].b,&p[i].rate);
         p[i].k=-p[i].a/p[i].b; p[i].id=i;
     }
     sort(p+,p+n+);
     solve(,n);
     printf("%.3lf",f[n]);
     return ;
 }