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Time Limit: 9000/3000 MS (Java/Others)    Memory Limit: 131072/65536 K (Java/Others)
Total Submission(s): 12824    Accepted Submission(s):
3967

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

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斜率优化dp学习:http://www.cnblogs.com/ka200812/archive/2012/08/03/2621345.html

#include<iostream>
#include<cstdio>
#include<cstring>

#define N 500005

using namespace std;
int dp[N],q[N],sum[N];
int head,tail,n,m;

int get_dp(int i,int j)
{
    return dp[j]+m+(sum[i]-sum[j])*(sum[i]-sum[j]);
}

int get_up(int j,int k)
{
    return dp[j]+sum[j]*sum[j]-(dp[k]+sum[k]*sum[k]);
}

int get_down(int j,int k)
{
    return 2*(sum[j]-sum[k]);
}

int main()
{
    while(scanf("%d%d",&n,&m)==2)
    {
        for(int i=1;i<=n;i++) scanf("%d",&sum[i]);
        sum[0]=dp[0]=0;head=tail=0;
        for(int i=1;i<=n;i++) sum[i]+=sum[i-1];
        q[tail++]=0;
        for(int i=1;i<=n;i++)
        {
            while(head+1<tail && get_up(q[head+1],q[head])<=sum[i]*get_down(q[head+1],q[head]))
              head++;
            dp[i]=get_dp(i,q[head]);
            while(head+1<tail && get_up(i,q[tail-1])*get_down(q[tail-1],q[tail-2])<=get_up(q[tail-1],q[tail-2])*get_down(i,q[tail-1]))
              tail--;
            q[tail++]=i;
        }
        printf("%d\n",dp[n]);
    }
    return 0;
}