hdu_3507_Print Article

Print Article

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

Total Submission(s): 8558    Accepted Submission(s): 2664


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

斜率dp入门题，第一次接触，学习一下，kuangbin大神写得很好，膜拜~~~~~通过队列来维护一个下凹的图形，斜率逐渐增大。

#include <iostream>
#include <cstdio>
#include <cstdlib>
#include <cstring>
#include <cmath>
#include <stack>
#include <bitset>
#include <queue>
#include <set>
#include <map>
#include <string>
#include <algorithm>
#define Si(a) scanf("%d",&a)
#define Sl(a) scanf("%lld",&a)
#define Sd(a) scanf("%lf",&a)
#define Ss(a) scanf("%s",a)
#define Pi(a) printf("%d\n",(a))
#define Pl(a) printf("%lld\n",(a))
#define Pd(a) printf("%lf\n",(a))
#define Ps(a) printf("%s\n",(a))
#define W(a) while(a--)
#define mem(a,b) memset(a,(b),sizeof(a))
#define FOP freopen("data.txt","r",stdin)
#define inf 0x3f3f3f3f
#define maxn 500010
#define mod 1000000007
#define PI acos(-1.0)
#define LL long long
using namespace std;

int n,m;

int sum[maxn];
int dp[maxn];
int q[maxn];//queue

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

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

int getDOWN(int j,int k)
{
return 2*(sum[j]-sum[k]);
}

int main()
{
int i,j;
while((scanf("%d%d",&n,&m))==2)
{
int val;
sum[0]=dp[0]=0;
for(i=1;i<=n;i++)
{
Si(val);
sum[i]=sum[i-1]+val;
}
int head=0,tail=0;
q[tail++]=0;
for(i=1;i<=n;i++)
{
while(head+1<tail&&getUP(q[head+1],q[head])<=getDOWN(q[head+1],q[head])*sum[i])
head++;
dp[i]=getDP(i,q[head]);
while(head+1<tail&&getUP(i,q[tail-1])*getDOWN(q[tail-1],q[tail-2])<=
getUP(q[tail-1],q[tail-2])*getDOWN(i,q[tail-1]))
tail--;
q[tail++]=i;
}
Pi(dp
);
}

return 0;
}