hdu 3507Print Article

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Time Limit: 9000/3000 MS (Java/Others)    Memory Limit: 131072/65536 K (Java/Others)
Total Submission(s): 5443    Accepted Submission(s): 1678

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 559575

 

Sample Output

230

 

这道题是斜率优化dp的典型题目，具体思路参考：点击打开链接

不过非常不幸，一开始我斜率是用double直接除了，一直WA。后来看了别人的代码，发现都是分 分子和分母写的，于是也改成这种风格，终于AC了。

看来这么一小点的误差都不容许啊，以后要小心了。注意：这道题分母是负数，分母乘过去时不等号要改变。

#include<cstdio>#include<iostream>#include<cstring>#define Maxn 500010using namespace std;int q[Maxn];long long sum[Maxn],dp[Maxn];long long fun_up(int a,int b){    return dp[a]-dp[b]+sum[a]*sum[a]-sum[b]*sum[b];}long long fun_dn(int a,int b){  //负数    return 2.0*(sum[a]-sum[b]);}int main(){    int n,m;    while(~scanf("%d%d",&n,&m)){        for(int i=1;i<=n;i++){            scanf("%I64d",sum+i);            sum[i]+=sum[i-1];        }        dp[0]=0,dp[1]=sum[1]*sum[1]+m;        int front=0,rear=1;        q[0]=0,q[1]=1;        for(int i=2;i<=n;i++){            while(front<rear&&fun_up(q[front],q[front+1])>=sum[i]*fun_dn(q[front],q[front+1]))                ++front;            int j=q[front];            dp[i]=dp[j]+(sum[i]-sum[j])*(sum[i]-sum[j])+m;            while(front<rear&&fun_up(q[rear-1],q[rear])*fun_dn(q[rear],i)>=fun_up(q[rear],i)*fun_dn(q[rear-1],q[rear]))                --rear;            q[++rear]=i;        }        printf("%I64d\n",dp[n]);    }return 0;}