DP(斜率优化)：HDU 3507 Print Article

Print Article

Time Limit: 9000/3000 MS (Java/Others)    Memory Limit: 131072/65536 K (Java/Others)
Total Submission(s): 8199    Accepted Submission(s): 2549

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

 

　　学到了，这样的斜率优化。

　　

#include <iostream>
#include <cstdio>
using namespace std;
const int maxn=500010;
long long f[maxn],s[maxn];
int q[maxn]; 
int main()
{
    int n,m,front,back;
    while(~scanf("%d%d",&n,&m))
    {
        s[0]=f[0]=0;
        for(int i=1;i<=n;i++)scanf("%d",&s[i]);
        for(int i=2;i<=n;i++)s[i]+=s[i-1];
        
        front=1;back=2;
        q[front]=0;
        
        for(int i=1;i<=n;i++)
        {
            while(front<back-1&&(f[q[front+1]]+s[q[front+1]]*s[q[front+1]])-(f[q[front]]+s[q[front]]*s[q[front]])<=2*s[i]*(s[q[front+1]]-s[q[front]])) front++;
            
            f[i]=f[q[front]]+(s[i]-s[q[front]])*(s[i]-s[q[front]])+m;
            
            while(front<back-1&&(s[q[back-1]]-s[q[back-2]])*((f[i]+s[i]*s[i])-(f[q[back-1]]+s[q[back-1]]*s[q[back-1]]))<=(s[i]-s[q[back-1]])*((f[q[back-1]]+s[q[back-1]]*s[q[back-1]])-(f[q[back-2]]+s[q[back-2]]*s[q[back-2]])))back--;
            q[back++]=i;
        }
        printf("%lld
",f[n]);
    }
    return 0;
}