[HDU 3507]Print Article（斜率优化Dp）

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.

Solution

斜率优化Dp入门题

代码明明很简单然而一开始写挂了（？？）我还对拍了啊喂…什么错都没拍出来

每次交都觉得能A，每次都证实是错觉，结果成为了HDU交的次数最多的一道题，想哭TvT

#include<iostream>
#include<cstdio>
#include<cstring>
#include<cstdlib>
#include<queue>
using namespace std;
int n,m;
int sum[500005],f[500005];
int q[1000005],head,tail;
int read()
{
    int x=0,f=1;char c=getchar();
    while(c<'0'||c>'9'){
        if(c=='-')f=-1;c=getchar();
    }
    while(c>='0'&&c<='9'){
        x=x*10+c-'0';c=getchar();
    }
    return x*f;
}
int getx(int i,int j)
{
    return f[j]+sum[j]*sum[j]-f[i]-sum[i]*sum[i];
}
int gety(int i,int j)
{
    return 2*(sum[j]-sum[i]);
}
int main()
{
    while(~scanf("%d%d",&n,&m))
    {
        for(int i=1;i<=n;i++)
        {
            sum[i]=read();
            sum[i]+=sum[i-1];
        }
        head=tail=0;q[tail++]=0;
        for(int i=1;i<=n;i++)
        {
            while(head+1<tail&&getx(q[head],q[head+1])<=sum[i]*gety(q[head],q[head+1]))//要取到等号
            head++;
            f[i]=f[q[head]]+(sum[i]-sum[q[head]])*(sum[i]-sum[q[head]])+m;
            while(head+1<tail&&getx(q[tail-1],i)*gety(q[tail-2],q[tail-1])<=getx(q[tail-2],q[tail-1])*gety(q[tail-1],i))//顺序不能反否则符号会变
            tail--;
            q[tail++]=i;
        }
        printf("%d
",f[n]);
    }
    return 0;
}