Print Article
Time Limit: 9000/3000 MS (Java/Others) Memory Limit: 131072/65536 K (Java/Others)
Total Submission(s): 8199 Accepted Submission(s): 2549
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.
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
5
9
5
7
5
Sample Output
230
学到了，这样的斜率优化。
#include <iostream>
#include <cstdio>
using namespace std;
const int maxn=;
long long f[maxn],s[maxn];
int q[maxn];
int main()
{
int n,m,front,back;
while(~scanf("%d%d",&n,&m))
{
s[]=f[]=;
for(int i=;i<=n;i++)scanf("%d",&s[i]);
for(int i=;i<=n;i++)s[i]+=s[i-]; front=;back=;
q[front]=; for(int i=;i<=n;i++)
{
while(front<back-&&(f[q[front+]]+s[q[front+]]*s[q[front+]])-(f[q[front]]+s[q[front]]*s[q[front]])<=*s[i]*(s[q[front+]]-s[q[front]])) front++; f[i]=f[q[front]]+(s[i]-s[q[front]])*(s[i]-s[q[front]])+m; while(front<back-&&(s[q[back-]]-s[q[back-]])*((f[i]+s[i]*s[i])-(f[q[back-]]+s[q[back-]]*s[q[back-]]))<=(s[i]-s[q[back-]])*((f[q[back-]]+s[q[back-]]*s[q[back-]])-(f[q[back-]]+s[q[back-]]*s[q[back-]])))back--;
q[back++]=i;
}
printf("%lld\n",f[n]);
}
return ;
}