笔记-[APIO2010]特别行动队
(f_i) 表示将 ((j+1,j+2,dots,i)) 分为一组，已解决 (i) 之前的士兵的最小代价。
(a<0)。
[egin{split}
f_i=&max{f_j+aX^2+bX+c}(j<i)\
=&max{f_j+aleft(sum_{h=j+1}^ix_h
ight)^2+bleft(sum_{h=j+1}^ix_h
ight)+c}\
end{split}
]
设 (s_i=sum_{h=1}^i x_h)：
[egin{split}
f_i=&max{f_j+aleft(sum_{h=j+1}^ix_h
ight)^2+bleft(sum_{h=j+1}^ix_h
ight)+c}\
=&max{f_j+a(s_i-s_j)^2+b(s_i-s_j)+c}\
=&max{f_j+a(s_i^2-2s_is_j+s_j^2)+b(s_i-s_j)+c}\
=&max{f_j+as_i^2-2as_is_j+as_j^2+bs_i-bs_j+c}\
=&max{f_j-2as_is_j+as_j^2-bs_j}+as_i^2+bs_i+c\
end{split}
]
考虑 (j=k) 比 (j=t) 更优：
[egin{split}
f_k-2as_is_k+as_k^2-bs_k>&f_t-2as_is_t+as_t^2-bs_t\
f_k-2as_is_k+as_k^2-bs_k>&f_t-2as_is_t+as_t^2-bs_t\
(f_k+as_k^2-bs_k)-(f_t+as_t^2-bs_t)>&2as_is_k-2as_is_t\
frac{(f_k+as_k^2-bs_k)-(f_t+as_t^2-bs_t)}{s_k-s_t}>&2as_i\
end{split}
]
搞定。
Code
#include <bits/stdc++.h>
using namespace std;
//Start
#define re register
#define il inline
#define mk make_pair
#define pb push_back
#define db double
#define lng long long
#define fi first
#define se second
const int inf=0x3f3f3f3f;
const lng INF=0x3f3f3f3f3f3f3f3f;
//Data
int n,a,b,c;
vector<int> x;
vector<lng> s,f;
//DP
template<typename T> il T p2(re T x){return x*x;}
il db fx(re int x){return s[x];}
il db fy(re int x){return f[x]+p2(s[x])*a-s[x]*b;}
il db slope(re int k,re int t){return (fy(k)-fy(t))/(fx(k)-fx(t));}
il lng DP(){
re int l=1,r=0; re vector<int> q(n+7); q[++r]=0;
for(re int i=1;i<=n;i++){
while(l<r&&slope(q[l],q[l+1])>=s[i]*2*a) l++;
f[i]=f[q[l]]+p2(s[i]-s[q[l]])*a+(s[i]-s[q[l]])*b+c;
while(l<r&&slope(q[r-1],q[r])<=slope(q[r],i)) r--;
q[++r]=i;
}
return f[n];
}
//Main
int main(){
scanf("%d%d%d%d",&n,&a,&b,&c);
x=vector<int>(n+7);
s=f=vector<lng>(n+7);
for(re int i=1;i<=n;i++)
scanf("%d",&x[i]),s[i]=s[i-1]+x[i];
printf("%lld
",DP());
return 0;
}
[Hugecolor{#ddd}{ exttt{---END---}}
]