Print Article

　　　　　　　　　　　　　　　　　　　　Time Limit: 9000/3000 MS (Java/Others)    Memory Limit: 131072/65536 K (Java/Others)
　　　　　　　　　　　　　　　　　　　　　　　　　　 Total Submission(s): 7960    Accepted Submission(s): 2465

Problem Description

Zero
has an old printer that doesn‘t work well sometimes. As it is antique,
he still like to use it to print articles. But it is too old to work for
a long time and it will certainly wear and tear, so Zero use a cost to
evaluate this degree.
One day Zero want to print an article which has
N words, and each word i has a cost Ci to be printed. Also, Zero know
that print k words in one line will cost

M is a const number.
Now Zero want to know the minimum cost in order to arrange the article perfectly.

Input

There are many test cases. For each test case, There are two numbers N and M in the first line (0 ≤ n ≤ 500000, 0 ≤ M ≤ 1000). Then, there are N numbers in the next 2 to N + 1 lines. Input are terminated by EOF.

Output

A single number, meaning the mininum cost to print the article.

Sample Input

5 5
5
9
5
7
5

Sample Output

230

Author

Xnozero

Source

2010 ACM-ICPC Multi-University Training Contest（7）——Host by HIT

【思路】

斜率优化。

设f[i],则转移式为f[i]=min{f[j]+(C[i]-C[j])^2+M}，1<=j<i

进一步得:f[i]=min{ (f[j]+C[j]^2-2*C[i]*C[j])+(C[i]^2+M) }

       设y(j)=f[j]+C[j]^2，a[i]=-*C[i]，x(j)=C(j)，则f[i]=min{y(j)+2*a[i]*x(j)}+C[i]^2+M

则要求min p=y+2ax , 单调队列维护下凸包。

【代码】

 1 #include<cstdio>
 2 #include<cstring>
 3 #include<iostream>
 4 using namespace std;
 5
 6 const int N = 500000+10;
 7
 8 struct point { int x,y;
 9 }q[N],now;
10 int L,R,n,m,C[N],f[N];
11 int cross(point a,point b,point c) {
12     return (b.x-a.x)*(c.y-a.y)-(b.y-a.y)*(c.x-a.x);
13 }
14 void read(int& x) {
15     char c=getchar(); while(!isdigit(c)) c=getchar();
16     x=0; while(isdigit(c)) x=x*10+c-‘0‘ , c=getchar();
17 }
18 int main() {
19     while(scanf("%d%d",&n,&m)==2) {
20         for(int i=1;i<=n;i++)
21             read(C[i]) , C[i]+=C[i-1];
22         L=R=0;
23         for(int i=1;i<=n;i++) {
24             while(L<R && q[L].y-2*C[i]*q[L].x>=q[L+1].y-2*C[i]*q[L+1].x) L++;
25             now.x=C[i];                                    //计算xi
26             now.y=q[L].y-2*C[i]*q[L].x+2*C[i]*C[i]+m;    //计算yi=f[i]+b[i]^2 = min p+a[i]^2+b[i]^2+M
27             while(L<R && cross(q[R-1],q[R],now)<=0) R--;
28             q[++R]=now;
29         }
30         printf("%d\n",q[R].y-C[n]*C[n]);
31     }
32     return 0;
33 }