Bzoj [NOI2007] Currency exchange Cash (CDQ Division + slope Optimization)

1492: [NOI2007] Currency exchange cash time limit: 5 Sec Memory Limit: MB
Submit: 2454 Solved: 1078
[Submit] [Status] [Discuss] Descriptioninput the first row of two positive integers n, S, respectively, indicating the number of days the small Y can predict and the initial amount of money. Next n lines, the K line three real AK, BK, Ratek, meaning as described in the title output has only a real maxprofit, indicating the maximum amount of money that can be obtained at the end of the nth day operation. The answer retains 3 decimal places. Sample Input3 100
1 1 1
1 2 2
2 2 3Sample Output
225.000


According to CDQ's thesis, N*LGN should be written. However, I forcibly scan the convex hull every time, the complexity of more than one LGN t_t.

#include <cstdio> #include <cmath> #include <queue> #include <stack> #include <string># include<cstring> #include <iostream> #include <map> #include <vector> #include <algorithm > #include <stdlib.h> #include <set> #include <ctime> #include <cmath> #include <iomanip >using namespace Std;typedef long long ll;const int mmax = 100010;const int inf = 0x3fffffff;const double EPS = 1e-8;d Ouble A[mmax],b[mmax],r[mmax];d ouble F[mmax];d ouble dp[mmax];int sgn (double x)//ûºåº¯êý{if (fabs (x) <eps) re    Turn 0; return x<0?-1:1;}    struct point{double x, y;    Point (Double x=0.0,double y=0.0): X (x), Y (y) {} void Read () {scanf ("%lf%lf", &x,&y); }};typedef Point Vector; Vector operator + (vector a,vector B) {return vector (A.X+B.X,A.Y+B.Y);} Vector operator-(point A,point B) {return Vector (A.X-B.X,A.Y-B.Y);} Vector operator * (vector a,double p) {return vector (a.x*p,a.y*p);} Vector operator/(Vector a,double p) {return Vector (a.x/p,a.y/p);} BOOL operator < (const point &a,const point &b) {return a.x<b.x | | (A.X==B.X&AMP;&AMP;A.Y&LT;B.Y);} ôëë㲿 Ödouble Dot (Vector a,vector B) {return a.x*b.x+a.y*b.y;} Double Cross (Vector a,vector B) {return a.x*b.y-a.y*b.x;} Double x (int i) {return dp[i]*r[i]/f[i];} Double y (int i) {return dp[i]/f[i];} BOOL CMP (int I,int j) {return a[i]*b[j]>a[j]*b[i];}    int Convexhull (point *p,int n,point *poly) {sort (p,p+n);    int m=0;        for (int i=n-1;i>=0;i--) {When (M>1 && Cross (poly[m-1]-poly[m-2],p[i]-poly[m-2)) <=0) m--;    Poly[m++]=p[i]; } return m;}        Point P[mmax],poly[mmax];int tmp[mmax];void CDQ (int l,int r) {if (l==r) {Dp[l]=max (dp[l],dp[l-1]);    return;    } int mid= (L+R) >>1;    CDQ (L,mid);    int cnt=0;    for (int i=l;i<=mid;i++) p[cnt++]=point (x (i), Y (i));    Cnt=convexhull (P,cnt,poly); for (int i=mid+1;i<=r;i++) tmp[i]=i;    Sort (tmp+mid+1,tmp+r+1,cmp);    int i=0,j=mid+1; while (J<=r) {while (i<cnt-1 && (POLY[I].Y-POLY[I+1].Y) *b[tmp[j]] < -1.0* (poly[i].x-poly[i+1].x        ) *a[tmp[j]) i++;        Dp[tmp[j]]=max (DP[TMP[J]],A[TMP[J]]*POLY[I].X+B[TMP[J]]*POLY[I].Y);    j + +; } CDQ (Mid+1,r);}    int main () {int n,s;    cin>>n>>s;        for (int i=1;i<=n;i++) {scanf ("%lf%lf%lf", &a[i],&b[i],&r[i]);    F[i]=a[i]*r[i]+b[i];    } memset (dp,0,sizeof DP);    dp[0]=1.0*s;    CDQ (1,n);    Double ans=0;    for (int i=1;i<=n;i++) Ans=max (Ans,dp[i]);    printf ("%.3lf\n", ans); return 0;}

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Bzoj [NOI2007] Currency exchange Cash (CDQ Division + slope Optimization)