hdu 3507 Print Article

Posted 2020-10-05 日拱一卒 功不唐捐

http://acm.hdu.edu.cn/showproblem.php?pid=3507

Time Limit: 9000/3000 MS (Java/Others)    Memory Limit: 131072/65536 K (Java/Others)

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

 

斜率优化

把斜率转化为乘积的形式，避免浮点数误差

#include<cstdio>
#include<cstring>
#define N 500001
using namespace std;
typedef long long LL;
LL sum[N],dp[N];
int c,q[N],head,tail; 
long long up(int k,int j) 
{
    return dp[j]-dp[k]-sum[k]*sum[k]+sum[j]*sum[j];
}
long long down(int k,int j)
{
    return 2*(sum[j]-sum[k]);
}
int main()
{
    int n,m,x;
    while(scanf("%d%d",&n,&m)!=EOF)
    {
        memset(sum,0,sizeof(sum));
        for(int i=1;i<=n;i++)
        {
            scanf("%d",&x);
            sum[i]=sum[i-1]+x;
        }
        head=tail=0;
        memset(q,0,sizeof(q));
        memset(dp,0,sizeof(dp));
        for(int i=1;i<=n;i++)
        {
            while(head<tail && up(q[head],q[head+1])<=sum[i]*down(q[head],q[head+1])) head++;
            dp[i]=dp[q[head]]+m+(sum[i]-sum[q[head]])*(sum[i]-sum[q[head]]);
            while(head<tail && up(q[tail-1],q[tail])*down(q[tail],i)>=up(q[tail],i)*down(q[tail-1],q[tail])) tail--;
            q[++tail]=i;
        }
        printf("%lld\n",dp[n]);
    }
}