hdu 3507 Print Article(DP+斜率优化)

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

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  

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;
}