hdu 3075 Print Article

题目描述：

Print Article

Time Limit: 9000/3000 MS (Java/Others)    Memory Limit: 131072/65536 K (Java/Others)
Total Submission(s): 2485    Accepted Submission(s): 815

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

第一道经典斜率dp！！！！

思路；
首先，这是一道dp ，很容易写出状态转移方程：
dp[i]=dp[j]+M+(sum[i]-sum[j])^2;

假设 有 k<j<i 且j决策更优 得到：
       dp[j]+(sum[i]-sum[j])^2<dp[k]+(sum[i]-sum[k])^2
化简：
       ((dp[j]+sum[j]^2-(dp[k]+sum[k]^2))<sum[i]*(sum[j]-sum[k])
设 y[i]=dp[i]+sum[i]^2 ,x[i]=sum[i]

上式等价于    (y[j]-y[k])/(x[j]-x[k])<sum[i];

用g(j,k) 表示上式左端 ;

下面证明 当k<j<i 时若g(j,k) >= g(i,j)那么j不能是最优解

1) : 若g(i,j) < sum[ii]  ,那么i比j优

2) : 若g(i,j)>=sum[ii],则g(j,k)>g(i,j)>sum[ii] k比j优

由上我们可以维护这么一个队列 （ 有人称之为凸性） 在队列中 若有k<j<i 则必有g(j,k)<g(i,j)

每次更新的时候，从队头开始查找，直到有其(y[he+1]-y(he])/(x[he+1]-x[he])<sum[i]) 为之 。

以上优化就称之斜率优化：

附上小代码：

#include<iostream>
#include<cstdio>
#include<cstring>
#include<algorithm>

#define inf 0x7777777
#define int long long

using namespace std;
int const nMax = 500010;

int a[nMax];
int dp[nMax];
int sum[nMax];
int n,M;

int q[nMax],he,ta;
int dy(int i,int j){
    return dp[i]+sum[i]*sum[i]-dp[j]-sum[j]*sum[j];
}
int dx(int i,int j){
    return 2*(sum[i]-sum[j]);
}
void query(int i){
    while(he<ta){
        if(dy(q[he+1],q[he])<=sum[i]*(dx(q[he+1],q[he])))he++;
        else break;
    }
    dp[i]=dp[q[he]]+(sum[i]-sum[q[he]])*(sum[i]-sum[q[he]])+M;
    return ;
}
void insert(int i){
    while(he<ta){
        int y2=q[ta],y1=q[ta-1];
        if(dy(y2,y1)*(dx(i,y2))>=dy(i,y2)*dx(y2,y1))ta--;
        else break;
    }
    q[++ta]=i;
    return ;
}

main()
{
    while(~scanf("%I64d%I64d",&n,&M)){
        sum[0]=0;
        for(int i=1;i<=n;i++){
            scanf("%I64d",&a[i]);
            sum[i]=sum[i-1]+a[i];
        }
        dp[0]=0;
        ta=he=0;
        q[0]=0;
        int ans=inf;
        for(int i=1;i<=n;i++){
            query(i);
            ans=min(ans,dp[i]);
            insert(i);
        }
        printf("%I64d\n",dp[n]);
    }
    return 0;
}