HDU-3507-Print Article-斜率优化-DP

最新推荐文章于 2021-03-16 21:24:00 发布

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本文介绍了一种通过斜率优化来解决最小成本打印问题的方法。该问题涉及如何以最低的成本安排文章的打印工作，考虑了每个单词的打印成本以及每行打印特定数量单词的额外固定成本。

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题目：

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

思路：简单的斜率优化

#include<stdio.h>
#include<iostream>
#include<string.h>
#include<queue>
using namespace std;
#define maxn 500050
int dp[maxn];
int gg[maxn];
int sum[maxn];
int n,m;
int getDP(int i,int j) {    //求出连续区间的一段值.即最基本的DP方法。
    return dp[j]+m+(sum[i]-sum[j])*(sum[i]-sum[j]);
}
int getUP(int j,int k) {    //斜率上部分
    return dp[j]+sum[j]*sum[j]-(dp[k]+sum[k]*sum[k]);
}
int getDOWN(int j,int k) {  //斜率下部分
    return 2*(sum[j]-sum[k]);
}
int main()
{
    while(scanf("%d%d",&n,&m)!=EOF) {
        memset(sum,0,sizeof sum);
        memset(dp,0,sizeof dp);
        for(int i=1;i<=n;i++)
           scanf("%d",&sum[i]);
        for(int i=1;i<=n;i++)
           sum[i]+=sum[i-1];
        int head=0,wei=1;
        gg[1]=0;
        for(int i=1;i<=n;i++) {
            while(head+1<wei&&getUP(gg[head+1],gg[head])<=sum[i]*getDOWN(gg[head+1],gg[head]))
               head++;
            //排除第一种情况
            dp[i]=getDP(i,gg[head]);
            while(head+1<wei&&getUP(i,gg[wei-1])*getDOWN(gg[wei-1],gg[wei-2])<=getUP(gg[wei-1],gg[wei-2])*getDOWN(i,gg[wei-1]))
                    wei--;
            //排除第二种情况
            gg[wei++]=i;
        }
        printf("%d\n",dp[n]);
    }
    return 0;
}