HDU3507 Print Article

Print Article

Time Limit: 9000/3000 MS (Java/Others)    Memory Limit: 131072/65536 K (Java/Others)

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

 

裸的斜率dp。。注意不等号方向。。样例数据挺好的。。不管你是什么符号都输出230..23333

#include<cstdio>
#include<cstdlib>
#include<iostream>
#include<algorithm>
using namespace std;
const int N = 500100;
#define For(i,n) for(int i=1;i<=n;i++)
#define Rep(i,l,r) for(int i=l;i<=r;i++)

int n,M,sum[N],dp[N],q[N];
//  dp[i] = max{dp[j] + (sj-si)^2 + M}
//-> dp[k] - dp[j] + sk^2 - sj^2 
//    --------------------------   > 2si 时 j比k优 
//              sk - sj

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

int Down(int i,int j){
    return sum[i]-sum[j]; 
}

void DP(){
    int l=0,r=1;
    For(i,n){
        while(l+1<r && Up(q[l],q[l+1])>=2*sum[i]*Down(q[l],q[l+1])) l++;
        dp[i]=dp[q[l]] + (sum[i]-sum[q[l]])*(sum[i]-sum[q[l]]) + M;
        while(l+1<r && Up(q[r-2],q[r-1])*Down(q[r-1],i)>=Up(q[r-1],i)*Down(q[r-2],q[r-1])) --r;
        q[r++]=i;
    } 
    printf("%d
",dp[n]);
}

int main(){
    while(scanf("%d%d",&n,&M)!=EOF){
        q[0]=dp[0]=0;
        For(i,n){
            scanf("%d",&sum[i]);
            sum[i]+=sum[i-1];
        }
        DP();
    }
    return 0;
}