hdu3507 Print Article 单调队列斜率优化DP

Print Article

Time Limit: 9000/3000 MS (Java/Others)    Memory Limit: 131072/65536 K (Java/Others)
Total Submission(s): 6694    Accepted Submission(s): 2065

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
   
   
   
   

  给出N和M，有N个单词，接下来N行为C[i](1<=i<=N)，连续打印一段的花费是这一段C的和的平方加上M，问打印整个文章的最小花费。

  dp[i]表示前i个单词的最小花费，sum[i]=c[1]+...+c[i]，于是dp[i]=max(dp[j]+(sum[i]-sum[j])^2+M)(j<i)，对于某个j，展开式子，dp[i]=dp[j]+sum[i]^2+sum[j]^2-2*sum[i]*sum[j]+M，对于另一个k，如果j比k优的话有dp[j]+sum[i]^2+sum[j]^2-2*sum[i]*sum[j]+M<=dp[k]+sum[i]^2+sum[k]^2-2*sum[i]*sum[k]+M，移项化简得(dp[j]+sum[j]^2-(dp[k]+sum[k]^2))/(2*(sum[j]-sum[k]))<=sum[i]，把设yj=dp[j]+sum[j]^2，xj=sum[j]，那么只需要维护一个斜率越来越大的曲线，也就是下凸曲线（如果上凸，三个点的话中间的点肯定不是最优），每次找到队列中斜率尽量大但是小于等于sum[i]的相邻的两个点，由此得到dp[i]，再把i点加入队列。由于sum[i]是逐渐增大的，所以从头到尾只用维护一遍这个队列。

#include<iostream>
#include<cstdio>
#include<string>
#include<cstring>
#include<vector>
#include<cmath>
#include<queue>
#include<stack>
#include<map>
#include<set>
#include<algorithm>
#pragma comment(linker, "/STACK:102400000,102400000")
using namespace std;

const int MAXN=500010;
const int MAXM=100010;

int N,M;
int dp[MAXN],q[MAXN],sum[MAXN];

int y(int j,int i){
    return dp[i]+sum[i]*sum[i]-(dp[j]+sum[j]*sum[j]);
}

int x(int j,int i){
    return 2*(sum[i]-sum[j]);
}

int main(){
    freopen("in.txt","r",stdin);
    while(scanf("%d%d",&N,&M)!=EOF){
        int t;
        sum[0]=0;
        for(int i=1;i<=N;i++){
            scanf("%d",&t);
            sum[i]=sum[i-1]+t;
        }
        dp[0]=0;
        int front=0,rear=0;
        q[0]=0;
        for(int i=1;i<=N;i++){
            while(front<rear&&y(q[front],q[front+1])<=x(q[front],q[front+1])*sum[i]) front++;
            dp[i]=dp[q[front]]+(sum[i]-sum[q[front]])*(sum[i]-sum[q[front]])+M;
            while(front<rear&&y(q[rear],i)*x(q[rear-1],q[rear])<=y(q[rear-1],q[rear])*x(q[rear],i)) rear--;
            q[++rear]=i;
        }
        printf("%d\n",dp[N]);
    }
    return 0;
}