HDU3507 Print Article

淡淡的烟草味﹌ 2022-07-21 11:11 21阅读 0赞

# Print Article #

**Time Limit: 9000/3000 MS (Java/Others) Memory Limit: 131072/65536 K (Java/Others)  
Total Submission(s): 9231 Accepted Submission(s): 2878**  
**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  
![3507-1.jpg][]  
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 ][2010 ACM-ICPC Multi-University Training Contest_7_Host by HIT]

**Recommend**

zhengfeng

此题是很基础的**斜率DP**的入门题。题意很清楚，就是输出序列a\[n\]，每连续输出的费用是连续输出的数字和的平方加上常数M让我们求这个费用的最小值。设dp\[i\]表示输出前i个的最小费用，那么有如下的DP方程：dp\[i\]= min\{ dp\[j\]+(sum\[i\]-sum\[j\])^2 +M \} 0<j<i

其中 sum\[i\]表示数字的前i项和。相信都能理解上面的方程。直接求解上面的方程的话复杂度是O(n^2)对于500000的规模显然是超时的。下面讲解下如何用斜率优化DP使得复杂度降低一维。

我们首先假设在算 dp\[i\]时，k<j ,j点比k点优。也就是dp\[j\]+(sum\[i\]-sum\[j\])^2+M <= dp\[k\]+(sum\[i\]-sum\[k\])^2+M;所谓j比k优就是DP方程里面的值更小.对上述方程进行整理很容易得到：\[(dp\[j\]+sum\[j\]\*sum\[j\])-(dp\[k\]+sum\[k\]\*sum\[k\])\] / 2(sum\[j\]-sum\[k\]) <=sum\[i\].

注意整理中要考虑下正负，涉及到不等号的方向。左边我们发现如果令：yj=dp\[j\]+sum\[j\]\*sum\[j\] xj=2\*sum\[j\]那么就变成了斜率表达式：（yj-yk）/(xj-xk) <= sum\[i\];而且不等式右边是递增的。

所以我们可以看出以下两点：我们令g\[k,j\]=(yj-yk)/(xj-xk)

第一：如果上面的不等式成立，那就说j比k优，而且随着i的增大上述不等式一定是成立的，也就是对i以后算DP值时，j都比k优。那么k就是可以淘汰的。

第二：如果 k<j<i 而且 g\[k,j\]>g\[j,i\] 那么 j 是可以淘汰的。假设 g\[j,i\]<sum\[i\]就是i比j优，那么j没有存在的价值.相反如果 g\[j,i\]>sum\[i\] 那么同样有 g\[k,j\]>sum\[i\] 那么 k比 j优 那么 j 是可以淘汰的

所以这样相当于在维护一个下凸的图形，斜率在逐渐增大。通过一个队列来维护。

int head,tail,n,m;
    // dp[i]= min{ dp[j]+M+(sum[i]-sum[j])^2 };
    int getDP(int i,int j)
    {
        return dp[j]+m+(sum[i]-sum[j])*(sum[i]-sum[j]);
    }
    
    int getUP(int j,int k) //yj-yk部分
    {
        return dp[j]+sum[j]*sum[j]-(dp[k]+sum[k]*sum[k]);
    }
    int getDOWN(int j,int  k)
    {
        return 2*(sum[j]-sum[k]);
    }
    
    int main()
    {
        while(scanf("%d%d",&n,&m)==2)
        {
            for(int i=1;i<=n;i++)
               scanf("%d",&sum[i]);
            sum[0]=dp[0]=0;
            for(int i=1;i<=n;i++)
               sum[i]+=sum[i-1];
            head=tail=0;
            q[tail++]=0;
            for(int i=1;i<=n;i++)
            {
                //把斜率转成相乘，注意顺序，否则不等号方向会改变的
                while(head+1<tail &&  getUP(q[head+1],q[head])<=sum[i]*getDOWN(q[head+1],q[head]))
                   head++;
                dp[i]=getDP(i,q[head]);
                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]))
                        tail--;
                q[tail++]=i;
            }
            printf("%d\n",dp[n]);
        }
        return 0;
    }

关于[程序算法艺术与实践][Link 1]更多讨论与交流，敬请关注本博客和新浪微博[songzi\_tea][songzi_tea].

[3507-1.jpg]: /images/20220720/ace37a6425ba48acada568bfaa1bd9fa.png
[2010 ACM-ICPC Multi-University Training Contest_7_Host by HIT]: http://acm.hdu.edu.cn/search.php?field=problem&key=2010+ACM-ICPC+Multi-University+Training+Contest%A3%A87%A3%A9%A1%AA%A1%AAHost+by+HIT+&source=1&searchmode=source
[Link 1]: http://blog.csdn.net/column/details/tac-programalgrithm.html
[songzi_tea]: http://weibo.com/songzitea