HDU 4258(Covered Walkway-斜率优化)
Covered Walkway
Time Limit: 30000/10000 MS (Java/Others) Memory Limit: 131072/131072 K (Java/Others)
Total Submission(s): 1273 Accepted Submission(s): 491
The building contractor has an interesting pricing scheme. To cover the walkway from a point atx to a point at
y, they will charge c+(x-y)2, wherec is a constant. Note that it is possible for
x=y. If so, then the contractor would simply chargec.
Given the points along the walkway and the constant c, what is the minimum cost to cover the walkway?
c (1≤c≤109), wheren is the number of points which must be covered, and
c is the contractor’s constant. Each of the followingn lines will contain a single integer, representing a point along the walkway that must be covered. The points will be in order, from smallest to largest. All of the points
will be in the range from 1 to 109, inclusive. The input will end with a line with two 0s.
yield answers which will fit in a signed 64-bit integer.
10 5000
1
23
45
67
101
124
560
789
990
1019
0 0
30726
pid=4257">4257
4260pid=4261">4261
4262设f[i]表示仅仅考虑前n个的最小费用
显然f[i]=f[j-1]+c+sqr(a[i]-a[j])
若选i比选j优(i<j)
f[i-1]+sqr(a[i]-a[y])+c<f[j-1]+sqr(a[j]-a[y])+c
f[i-1]+sqr(a[i])-f[j-1]-sqr(a[j])
----------------------------------- > 2*a[y]
a[i]-a[j]
斜率优化。
#include<cstdio>
#include<cstring>
#include<cstdlib>
#include<algorithm>
#include<functional>
#include<iostream>
#include<cmath>
#include<cctype>
#include<ctime>
using namespace std;
#define For(i,n) for(int i=1;i<=n;i++)
#define Fork(i,k,n) for(int i=k;i<=n;i++)
#define Rep(i,n) for(int i=0;i<n;i++)
#define ForD(i,n) for(int i=n;i;i--)
#define RepD(i,n) for(int i=n;i>=0;i--)
#define Forp(x) for(int p=pre[x];p;p=next[p])
#define Lson (x<<1)
#define Rson ((x<<1)+1)
#define MEM(a) memset(a,0,sizeof(a));
#define MEMI(a) memset(a,127,sizeof(a));
#define MEMi(a) memset(a,128,sizeof(a));
#define INF (2139062143)
#define F (100000007)
#define MAXN (1000000+10)
#define eps 1e-13
#define Read(x) { \
while (!isdigit(c=getchar())); \
x=c-48; \
while (isdigit(c=getchar())) x=x*10+c-48; \
}
long long mul(long long a,long long b){return (a*b)%F;}
long long add(long long a,long long b){return (a+b)%F;}
long long sub(long long a,long long b){return (a-b+(a-b)/F*F+F)%F;}
typedef long long ll;
int n;
ll cost;
char c;
ll a[MAXN],f[MAXN];
struct P
{
int i;
long double x,y;
P(int _i,ll _x,ll _y):i(_i),x(_x),y(_y){}
P(ll _x,ll _y):x(_x),y(_y){}
P(){}
friend long double kk(P a,P b){if (abs(a.x-b.x)<eps) return (b.y-a.y)*INF;return (b.y-a.y)/(b.x-a.x); }
}st[MAXN];
struct V
{
long double x,y;
V(ll _x,ll _y):x(_x),y(_y){}
V(){}
V(P a,P b):x(b.x-a.x),y(b.y-a.y){}
friend long double operator*(V a,V b){return a.x*b.y-a.y*b.x; }
};
int main()
{
freopen("B.in","r",stdin); while (1)
{
scanf("%d%lld",&n,&cost);
if (!n) break;
For(i,n)
Read(a[i]) f[0]=0;
int head=1,tail=1;
st[1]=P(1,a[1],a[1]*a[1]);
Fork(i,1,n)
{
P A=P(i,a[i],f[i-1]+a[i]*a[i]);
while (head^tail&&V(st[tail-1],st[tail])*V(st[tail],A)<=0) tail--;
st[++tail]=A; while (head^tail&&kk(st[head],st[head+1])<2*a[i]) head++;
int k=st[head].i;
f[i]=f[k-1]+cost+(a[k]-a[i])*(a[k]-a[i]);
}
cout<<f[n]<<endl; }
return 0;
}
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