Print Article[HDU 3507，2010 ACM-ICPC Multi-University Training Contest]

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Print Article

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

Solution

设 fi 表示前 i 个字母打出来的最少费用。

fi=mini−1j=0{fj+(∑k=j+1iCk)2+M}

推出斜率公式，A(i)⋅[2A(j)−2A(k)]>fj−fk+A2(j)−A2(k)

其中，A(x)=∑i=1xCi

Code

#include <iostream>
#include <cstdio>

#define LL long long

#define Min(x,y) ((x)<(y)?(x):(y))

using namespace std;

LL n,m,Max_Right,lastcost;
LL c[500010],A[500010],f[500010],As[500010];
double ks[500010];

int main(){

while(scanf("%lld%lld",&n,&m)!=EOF){

A[0]=0;
for(LL i=1;i<=n;i++){
scanf("%lld",&c[i]);
A[i]=A[i-1]+c[i];
}

lastcost=Max_Right=As[0]=f[0]=0;
ks[Max_Right+1]=9223372036854775807LL;ks[Max_Right]=-9223372036854775807LL;

for(LL i=1;i<=n;i++){
for(LL j=lastcost;j<=Max_Right;j++)
if(ks[j]<=A[i]&&ks[j+1]>=A[i]){
lastcost=j;
f[i]=f[j]+(A[i]-As[j])*(A[i]-As[j])+m;
for(LL k=Max_Right;k>=0;k--)
if((double)(f[i]+A[i]*A[i]-f[k]-As[k]*As[k])/(2*A[i]-2*As[k])>=ks[k]){
Max_Right=k+1;As[Max_Right]=A[i];f[Max_Right]=f[i];
ks[Max_Right]=(double)(f[i]+A[i]*A[i]-f[k]-As[k]*As[k])/(2*A[i]-2*As[k]);
ks[Max_Right+1]=9223372036854775807LL;
lastcost=Min(lastcost,Max_Right);
break;
}
break;
}
}

printf("%lld\n",f
);
}

return 0;
}