BZOJ1492: [NOI2007] Currency exchange cash

Title: http://www.lydsy.com/JudgeOnline/problem.php?id=1492

F[i]=max (F[i-1],x[j]*a[i]+y[j]*b[i]) F[i] represents the maximum profit, X[j],y[j] represents the maximum number of A/b coupons that can be swapped in J days respectively.

In slope optimization, if the slope of each state is monotonous, the x-coordinate of the point on the convex hull, the y-coordinate is monotonous. That last monotonous queue would be all right.

The slope given in this question and the x-coordinate on the convex hull are not monotonous ...

(Because the balance tree will not be written to maintain dynamic convex hull so wrote CDQ division.)

In fact... It's better to look at the paper honestly.

I'll just say a few words. Tat. The points that are inserted first contribute to the following points: If the point x, y coordinates on the convex hull can be ordered, the interval slope of the inquiry is orderly. Then you can maintain a stack ah or something.

So first to the slope of the order, divided by time into the left and right two intervals, the left side of the violence to make a convex bag, to ask questions. Finally, the two intervals are combined by x, Y (so the x, y coordinates are spicy).

#include <cstring>#include<iostream>#include<cstdio>#include<algorithm>#include<queue>#include<map>#include<cmath>#defineRep (i,l,r) for (int i=l;i<=r;i++)#defineDown (i,l,r) for (int i=l;i>=r;i--)#defineCLR (x, y) memset (x,y,sizeof (×))#defineMAXN 100500#defineINF Int (1E9)#definell Long Long#defineEPS 1e-9using namespacestd;structdata{DoubleK,a,b,x,y,r;intID;} A[MAXN],T[MAXN];intN,S[MAXN];DoubleF[MAXN];intRead () {intx=0, f=1;CharCh=GetChar ();  while(!isdigit (CH)) {if(ch=='-') f=-1; Ch=GetChar ();}  while(IsDigit (CH)) {x=x*Ten+ch-'0'; Ch=GetChar ();} returnx*F;} BOOLCMP (data a,data b) {returnA.k>B.K;}DoubleGETK (intAintB) {    if(! Breturn-1e20; if(Fabs (a[a].x-a[b].x) <eps)return1e20; return(A[B].Y-A[A].Y)/(a[b].x-a[a].x);}intCross (Data a,data b,data c) {Doublex1=b.x-a.x,y1=b.y-a.y,x2=c.x-a.x,y2=c.y-a.y; if(Fabs (x1*y2-x2*y1) <eps)return 0; if(x1*y2-x2*y1>0)return 1; return-1;}voidCdqintLintR) {    if(l==R) {F[l]=max (f[l],f[l-1]); A[l].y=f[l]/(a[l].a*a[l].r+a[l].b); a[l].x=a[l].r*a[l].y; return; }    intL1=l,l2= (L+R)/2+1, mid= (L+R)/2, j=1; Rep (I,l,r)if(A[i].id<=mid) t[l1++]=a[i];Elset[l2++]=A[i]; Rep (i,l,r) a[i]=T[i];    CDQ (L,mid); inttop=0; Rep (i,l,mid) { while(top>1&&cross (a[s[top-1]],a[s[top]],a[i]) >=0) top--; s[++top]=i; } s[++top]=0; Rep (I,mid+1, R) {         while(J&LT;TOP&AMP;&AMP;GETK (s[j],s[j+1]) +EPS&GT;A[I].K) J + +; intnow=a[i].id; F[now]=max (f[now],a[i].a*a[s[j]].x+a[i].b*a[s[j]].y); } CDQ (Mid+1, R); L1=l,l2=mid+1; Rep (I,l,r)if(l1<=mid&& (a[l1].x<a[l2].x| | (Fabs (a[l1].x-a[l2].x) <eps&&a[l1].y<a[l2].y) | | L2&GT;R)) t[i]=a[l1++];Elset[i]=a[l2++]; Rep (i,l,r) a[i]=t[i];}intMain () {n=read (); scanf"%LF", &f[0]); Rep (I,1, N) {scanf ("%LF%LF%LF", &AMP;A[I].A,&AMP;A[I].B,&AMP;A[I].R); A[i].id=i; A[I].K=-a[i].a/a[i].b; } sort (A+1, A +1+n,cmp); CDQ (1, N); printf ("%.3lf\n", F[n]); return 0;}

BZOJ1492: [NOI2007] Currency exchange cash