hdu 3507 斜率优化DP入门题

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

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
   
   
   
   

/**
hdu3507 斜率优化dp   
膜拜大神：http://www.cnblogs.com/ka200812/archive/2012/08/03/2621345.html
*/

#include <stdio.h>
#include <string.h>
#include <algorithm>
#include <iostream>
using namespace std;

const int N=500005;
int dp[N];
int q[N];
int 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]-2*sum[k];
}

int main()
{
    while(~scanf("%d%d",&n,&m))
    {
        for(int i=1; i<=n; i++)
            scanf("%d",&sum[i]);
        sum[0]=dp[0]=0;
        for(int i=1; i<=n; i++)
            sum[i]+=sum[i-1];
        head=tail=0;
        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;
}