#10191. 「一本通 5.6 练习 4」打印文章
这题™是多组数据。。。还搞了半天。。。
最后只好这样打才过了。。。
while (scanf("%d%d",&n,&m)==2)
{
}
斜率DP,式子很容易推:
f[i]=min{f[j]+(sum[i]-sum[j])2+a[i]2};
其中sum[]表示a[]的前缀和
上标:
#include<cstdio>
#include<algorithm>
#define N 500010
#define db double
#define ll long long
using namespace std;
int n,m,g[N],l,len;
ll a[N],sum[N],f[N];
inline int read()
{
int x=0; char c=getchar();
while (c<'0' || c>'9') c=getchar();
while (c>='0' && c<='9') x=(x<<1)+(x<<3)+(c^48),c=getchar();
return x;
}
ll sqr(ll x) {return x*x;}
ll solve(int x,int y) {return (db)(f[x]+sqr(sum[x])-f[y]-sqr(sum[y]))/(sum[x]-sum[y]);}
int main()
{
freopen("#10191.in","r",stdin);
freopen("#10191.out","w",stdout);
// f[i]=min{f[j]+(sum[i]-sum[j])^2+a[i]^2};
while (scanf("%d%d",&n,&m)==2)
{
for (int i=1;i<=n;i++)
a[i]=read(),sum[i]=sum[i-1]+a[i];
l=len=0;
for (int i=1;i<=n;i++)
{
while (l<len && solve(g[l+1],g[l])<2*sum[i]) l++;
f[i]=f[g[l]]+sqr(sum[i]-sum[g[l]])+m;
while (l<len && solve(g[len],g[len-1])>solve(i,g[len])) len--;
g[++len]=i;
}
printf("%lld\n",f[n]);
}
return 0;
}
转载需注明出处。