f[i]表示a[1]+…+a[i]+i。为了方便处理。。。dp[i]表示把前i件压缩需要的最少费用。则 dp[i]=min{dp[j]+(f[i]−f[j]−c)2|1≤j<i} 若k1< k2且k1优于k2，则: (dp[k2]+(f[k2]+c)2−dp[k1]−(f[k1]+c)2)/(2∗(f[k2]−f[k1]))>f[i] 。依然是维护一个下凸曲线，保证队列中的斜率是单增的。
#include <cstdio> #include <cstring> #define ll long long int const N=50000+10; ll dp[N],f[N]; int n,l,q[N],h=0,t=0,c=0; inline int read(){ int x=0;char ch=getchar(); while(ch<'0'||ch>'9') ch=getchar(); while(ch>='0'&&ch<='9') x=x*10+ch-'0',ch=getchar(); return x; } inline ll sqr(ll x){return x*x;} inline double slope(int k1,int k2){ return (dp[k2]-dp[k1]+sqr(f[k2]+c)-sqr(f[k1]+c))*1.0/(2.0*(f[k2]-f[k1])); } int main(){ // freopen("a.in","r",stdin); n=read();l=read();c=l+1; for(int i=1;i<=n;++i) f[i]=read(),f[i]+=f[i-1]+1; for(int i=1;i<=n;++i){ while(h<t&&slope(q[h],q[h+1])<f[i]) ++h; dp[i]=dp[q[h]]+sqr((f[i]-f[q[h]]-c)); while(h<t&&slope(q[t],i)<slope(q[t-1],q[t])) --t; q[++t]=i; } printf("%lld",dp[n]); return 0; }