Bzoj 1010: [HNOI2008] Toy packing toy--slope optimization

Bzoj 1010

The second slope optimization has a new understanding of the slope optimization. Happy (Foggy

The question is actually about slope optimization. Slope optimization. Slope optimization.

Okay, that's a naked question.

The node that we selected when we considered updating f[i]

If K is more superior than J

Then there are:

f[j]+ (j-i+sum[i]-sum[j]-l) ^2<f[k]+ (l+k-i+sum[i]-sum[k]-l) ^2

f[j]+ (Sum[i]-sum[j]) ^2-2* (i-j+l) * (Sum[i]-sum[j]) + (i-j+l) ^2<f[k]+ (Sum[i]-sum[k]) ^2+2* (i-k+l) * (Sum[i]-sum[k]) + (I-K+L) ^2

Make t1=i-j+l,t2=i-k+l;

f[j]+ (Sum[i]-sum[j]) ^2-2*t1* (Sum[i]-sum[j]) +t1^2<f[k]+ (Sum[i]-sum[k]) ^2-2*t2* (Sum[i]-sum[k]) +t2^2

f[j]-f[k]+ (Sum[i]-sum[j]) ^2-(Sum[i]-sum[k]) ^2+t1^2-t2^2<2*sum[i]* (SUM[K]-SUM[J]+K-J)

f[j]-f[k]+ (Sum[i]-sum[j]) ^2-(Sum[i]-sum[k]) ^2+t1^2-t2^2)/(2*SUM[K]-SUM[J]+K-J) <sum[i]

。。 Although this formula is very annoying, in fact, if you use Sum[i] to express sum[i]+i words is very convenient, but here I wrote a strict point

Well, Sum[i], the formula is as follows

(f[j]-f[k]+ (sum[j]+l) ^2+ (sum[k]+l) ^2)/((Sum[k]-sum[j]) <sum[i]

Okay, that's it.

Considering this thing is a slope equation, we ask for the minimum value, so the slope cannot be decremented.

Why. You might as well just update from the previous location because you're descending.

So we can maintain a lower convex hull according to this nature.

By the way, if the maximum value is to be maintained, the upper convex hull

Tween

Suddenly found that forgot to put the code ....

#include <iostream> #include <cstdio> #include <cmath> #include <algorithm> #include <queue&
Gt #include <string> #include <map> #include <set> #include <cstring> #include <ctime> # Include <vector> #define INF 1e9 #define EPS 1e-9 #define ITER multiset<ll>::iterator #define LL Long #d Efine MAXN 20010 #define for (i,j,k) for (ll i=j;i<=k;i++) #define DOW (I,j,k) for (ll i=k;i>=j;i--) using namespace std
;   inline void Read (ll &tx) {ll x=0,f=1;char Ch=getchar (); while (ch< ' 0 ' | |  Ch> ' 9 ') {if (ch== '-') F=-1;ch=getchar ();}  while (ch>= ' 0 ' &&ch<= ' 9 ') {x=x*10+ch-' 0 '; Ch=getchar ();} Tx=x*f;   } inline void Write (ll x) {if (x<0) Putchar ('-'), x=-x; if (x>=10) write (X/10);  Putchar (x%10+ ' 0 ');
} inline void Writeln (ll x) {write (x);p UTS ("");
ll n,l,a[100001],q[100001],l,r,f[100001]; inline ll Sqr (ll x) {return x*x;} inline double Get (ll y,ll x) {return (double) (F[X]-F[Y]+SQR (a[x]+l)-SQR (a[y]+l))/(2.0* (A[x]-a[y]));
	} int main () {read (n); Read (l); l++;
	for (i,1,n) read (A[i]);
	for (i,1,n) a[i]+=a[i-1]+1;
	l=r=1;q[1]=0;
		for (I,1,n) {while (L<r&&get (q[l],q[l+1]) <=a[i]) l++;
		F[I]=F[Q[L]]+SQR (A[I]-A[Q[L]]-L);
		while (L<r&&get (Q[r-1],q[r]) >get (q[r],i)) r--;
	Q[++r]=i;
} writeln (F[n]);	 }