Hdu 3507 Print Article (线性DP)

最新推荐文章于 2020-05-21 21:25:44 发布

原创最新推荐文章于 2020-05-21 21:25:44 发布 · 930 阅读

0 ·

CC 4.0 BY-SA版权

暑期训练（一）线性dp 专栏收录该内容

2 篇文章

订阅专栏

本文介绍了一种通过动态规划解决的算法问题，旨在寻找打印一篇包含特定数量单词的文章的最小成本。文章详细解释了问题背景及求解过程，并提供了一个具体的代码实现案例。

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

Sample Output

Author

Xnozero

题意略；

思路：本题是比较基本的线性规划

状态转移方程为: dp[i] = min(dp[j] + (sum[i]-sum[j])^2 + m) (j < i);

性质类似于四边形不等式

#include <cstdio>
#include <cstring>
#include <iostream>
using namespace std;
#define hehe cout<<"shabi"<<endl

long long dp[500010],sum[500010];

int m,n,s[500010],date;
void dpe()
{
    long long temp;
    dp[0]=0;
    s[0]=0;
    for (int i=1;i<=n;i++)
    {
        dp[i]=1e18;
        for (int j=s[i-1];j<=i;j++)
        {
            temp=dp[j]+(sum[i]-sum[j])*(sum[i]-sum[j])+m;
            if(dp[i]>=temp)
            {
                s[i]=j;
                dp[i]=temp;
            }
        }
    }
}
int main ()
{
    //freopen ("out.txt","r",stdin);
    while (scanf("%d%d",&n,&m)!=EOF)
    {
        memset(dp,0,sizeof(dp));
        memset(sum,0,sizeof(sum));
        memset(s,0,sizeof(s));
        for (int i=1;i<=n;i++) {
            scanf ("%d",&date);
            sum[i]=sum[i-1]+date;
        }
        dpe();
        printf("%lld\n",dp[n]);
    }
    
}