bzoj 3156

Posted 2020-06-11

f[i]=max{f[j]+(i-j-1)*i-(b[i-1]-b[j])+a[i]}b[i]为i的前缀和

易得(f[j]+b[j]-f[k]-b[k])/(j-k)<i

同样单调队列维护凸包

longlong老是没注意，AC率就是这么刷下来的QAQ

 1 #include<bits/stdc++.h>
 2 #define inc(i,l,r) for(int i=l;i<=r;i++)
 3 #define dec(i,l,r) for(int i=l;i>=r;i--)
 4 #define link(x) for(edge *j=h[x];j;j=j->next)
 5 #define mem(a) memset(a,0,sizeof(a))
 6 #define inf 1e9
 7 #define ll long long
 8 #define succ(x) (1<<x)
 9 #define NM 1000000+5
10 using namespace std;
11 int read(){
12     int x=0,f=1;char ch=getchar();
13     while(!isdigit(ch)){if(ch==‘-‘)f=-1;ch=getchar();}
14     while(isdigit(ch))x=x*10+ch-‘0‘,ch=getchar();
15     return x*f;
16 }
17 int n,a[NM],q[NM],qh,qt;
18 ll f[NM],b[NM];
19 double slope(int x,int y){
20     return (double)(f[y]+b[y]-f[x]-b[x])/(y-x);
21 }
22 int main(){
23     freopen("data.in","r",stdin);
24     n=read();
25     inc(i,1,n)a[i]=read();
26     inc(i,1,n)b[i]=b[i-1]+i;
27 //    inc(i,1,n)printf("%d ",b[i]);printf("\n");
28     qh=qt=1;q[1]=0;
29     inc(i,1,n){
30         while(qh+1<=qt&&slope(q[qh],q[qh+1])<i)qh++;
31         int j=q[qh];
32         f[i]=f[j]+(ll)(i-j-1)*i-b[i-1]+b[j]+a[i];
33         while(qh+1<=qt&&slope(q[qt-1],q[qt])>=slope(q[qt],i))qt--;
34         q[++qt]=i;
35     }
36 //    inc(i,1,n)printf("%d ",f[i]);printf("\n");
37     printf("%lld\n",f[n]);
38     return 0;
39 }

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