hdu 3507 Print Article(DP+斜率优化)

最新推荐文章于 2019-06-24 05:57:30 发布

原创最新推荐文章于 2019-06-24 05:57:30 发布 · 1.2k 阅读

1 ·

CC 4.0 BY-SA版权

文章标签：

#优化 #numbers #training #input #each #output

DP 专栏收录该内容

97 篇文章

订阅专栏

本文探讨了一种古老打印机在排版文章时的成本优化策略，通过数学模型计算出最小花费的方法。针对给定的单词数量和成本函数，作者提出了高效的算法来确定最优的排版方案，确保文章的完美打印同时降低成本。

Print Article

Time Limit: 9000/3000 MS (Java/Others) Memory Limit: 131072/65536 K (Java/Others)
Total Submission(s): 2390 Accepted Submission(s): 790

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

Source

2010 ACM-ICPC Multi-University Training Contest（7）——Host by HIT

Recommend

zhengfeng

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

题意：给你n个单词，要你决定一个排版的策略使得花费最小，花费的公式在上面

分析：很容易的一道题，写出简单的转移

f[ i ]=min{ f[ j-1 ] +w[ j , i ] + m} (1<=j<=i)

w[ j , i ]=sum{ c[ j ] + c[ j+1 ] +... +c[ i ] }

设：s[ i ]=sum{ c[ 1 ] + c[ 2 ]+... + c[ i ] }

综上可以得到

f[ i ]=min{ f[ j-1 ]+s[ j-1 ]^2 -2 s[ i ]*s[ j-1 ] }+s[ i ]^2+m

令 a=2s[ i ] x=s[ j- 1] y= f[ j-1 ]+s[ j-1 ]^2

G=-ax+y

即y=ax+G，很明显的斜率优化了。。。

这题丝毫没有坑人的地方吧，1Y了

代码：

#include<cstdio>
#include<iostream>
using namespace std;
const int mm=555555;
int f[mm],s[mm],q[mm];
int i,j,n,m,l,r;
bool TRight(int ax,int ay,int bx,int by,int cx,int cy)
{
    return (ax-bx)*(cy-by)>=(ay-by)*(cx-bx);
}
int gx(int i)
{
    return s[i-1];
}
int gy(int i)
{
    return f[i-1]+s[i-1]*s[i-1];
}
int get(int i,int a)
{
    return gy(i)-a*gx(i);
}
int main()
{
    while(~scanf("%d%d",&n,&m))
    {
        f[0]=s[0]=l=0,r=-1;
        for(i=1;i<=n;++i)
        {
            scanf("%d",&s[i]);
            s[i]+=s[i-1];
        }
        for(i=1;i<=n;++i)
        {
            while(l<r&&TRight(gx(q[r-1]),gy(q[r-1]),gx(q[r]),gy(q[r]),gx(i),gy(i)))--r;
            q[++r]=i;
            while(l<r&&get(q[l],2*s[i])>=get(q[l+1],2*s[i]))++l;
            f[i]=get(q[l],2*s[i])+m+s[i]*s[i];
        }
        printf("%d\n",f[n]);
    }
    return 0;
}