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于 2019-06-24 05:57:30 发布

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原文链接：http://www.cnblogs.com/sdau--codeants/p/3251238.html

博客围绕文章打印最小成本问题展开，给出输入输出要求及示例。还提及斜率优化，给出相关链接。同时展示了求解该问题的代码，包括初始化、求dp值、分子分母等函数及主函数逻辑，通过一系列操作计算并输出最小成本。

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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

Author

Xnozero

Source

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

Recommend

zhengfeng

***************************************************************************************

斜率优化链接： http://www.cnblogs.com/ka200812/archive/2012/08/03/2621345.html

讲解的很清楚，有图又推导，不过看了很长时间才看懂，关键求斜率推导式

***************************************************************************************

 1 #include<iostream>
 2 #include<cstring>
 3 #include<string>
 4 #include<cstdio>
 5 #include<cmath>
 6 #define  L  long long
 7 using namespace std;
 8 L dp[500005];
 9 L sum[500005];
10 int  q[500005];
11 int n,m,i,j;
12 void  init()
13 {
14     for(i=1;i<=n;i++)
15      cin>>sum[i];
16     for(i=2;i<=n;i++)
17       sum[i]+=sum[i-1];
18 
19 }
20 L getdp(int i,int j)//求dp
21     {
22         return dp[j]+m+(sum[i]-sum[j])*(sum[i]-sum[j]);
23     }
24 L  getup(int j,int k)//分子
25  {
26      return   (dp[j]+sum[j]*sum[j])-(dp[k]+sum[k]*sum[k]);
27  }
28 L getdown(int j,int k)//分母
29   {
30       return  2*(sum[j]-sum[k]);
31   }
32 int main()
33 {
34    while(scanf("%d %d",&n,&m)!=EOF)
35     {
36         memset(dp,0,sizeof(dp));
37         memset(sum,0,sizeof(sum));
38         memset(q,0,sizeof(q));
39         init();
40       int head=0;
41     int tail=0;
42      q[0]=dp[0]=0;
43      tail++;
44      for(i=1;i<=n;i++)
45       {
46           while(head+1<tail&&getup(q[head+1],q[head])<=sum[i]*getdown(q[head+1],q[head]))
47             head++;
48             dp[i]=getdp(i,q[head]);
49           while(head+1<tail&&getup(i,q[tail-1])*getdown(q[tail-1],q[tail-2])<=getup(q[tail-1],q[tail-2])*getdown(i,q[tail-1]))//删去一定不满足条件的点
50             tail--;
51           q[tail++]=i;
52       }
53 
54   cout<<dp[n]<<endl;
55 }
56   return 0;
57 
58 
59 
60 }