Print Article HDU - 3507

最新推荐文章于 2021-03-16 21:24:00 发布

转载最新推荐文章于 2021-03-16 21:24:00 发布 · 160 阅读

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原文链接：http://www.cnblogs.com/plysc/p/10774407.html

本文介绍了一道关于最小化打印成本的斜率DP题目，通过动态规划算法，在考虑每行字数成本的基础上，求解打印一篇包含N个单词的文章的最低总成本。输入包括单词数量N和每行固定成本M，以及每个单词的打印成本，输出为最小打印成本。

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一道斜率dp入门题

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.

InputThere 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.OutputA single number, meaning the mininum cost to print the article.Sample Input

Sample Output

#include<cstdio>
#include<iostream>
#include<cstring>
#include<algorithm>
#define ll long long
using namespace std;
const int maxn = 1e6+10;

int n,tail,head,q[maxn],m;

ll f[maxn],sum[maxn];

double g(int k2,int k1) {
    return double(f[k1]-f[k2]+sum[k1]*sum[k1]-sum[k2]*sum[k2])/double(2*sum[k1]-2*sum[k2]);
}

int main() {
    while(scanf("%d%d",&n,&m)!=EOF){
        for(int i=1;i<=n;i++) {
            scanf("%lld",&sum[i]);sum[i]+=sum[i-1];
            if(sum[i]==sum[i-1]) i--,n--;//不加会错。。
        }
        head=tail=1;f[0]=0;
        for(int i=1;i<=n;i++) {
            int k1=q[head],k2=q[head+1];
            while(head<tail&&g(k1,k2)<=sum[i]) 
            head++,k1=q[head],k2=q[head+1];
            f[i]=f[k1]+(sum[i]-sum[k1])*(sum[i]-sum[k1])+m;
            while(head<tail&&g(q[tail-1],q[tail])>=g(q[tail],i)) tail--;
            q[++tail]=i;
        }
        printf("%lld\n",f[n]);
    }
    return 0;
}